Can the Stochastic Wave Equation with Strong Drift Hit Zero?

We study the stochastic wave equation with multiplicative noise and singular drift: \[ \partial_tu(t,x)=\Delta u(t,x)+u^{-\alpha}(t,x)+g(u(t,x))\dot{W}(t,x) \] where $x$ lies in the circle $\mathbf{R}/J\mathbf{Z}$ and $u(0,x)>0$. We show that (i) If $0<\alpha<1$ then with positive probability, $u(t,x)=0$ for some $(t,x)$. (ii) If $\alpha>3$ then with probability one, $u(t,x)\ne0$ for all $(t,x)$.


Introduction
One of the classic questions about stochastic processes is whether they can hit a given set. That is, for a process X t taking values in a space S, and for A ⊂ S, do we have P(X t ∈ A for some t) > 0.
For example, consider the Bessel process R t with parameter n, which satisfies dR = n − 1 2R dt + dW where W (t) is a one-dimensional Brownian motion and we assume that R 0 > 0. It is well known that if we allow n to take nonnegative real values, then R t can hit 0 iff n < 2. For Markov processes such as R t , harmonic functions and potential theory are powerful tools which have led to rather complete answers to such questions; see [MP10] or most other books in Markov processes. For stochastic partial differential equations (SPDE), potential theory becomes less tractible due to the infinite-dimensional state space of solutions, and hitting questions have not been as thoroughly studied. To be specific, solutions u(t, x) usually depend on a time parameter t 1 and a spatial parameter x. So for a fixed time t, the solution u(t, x) is a function of x, and the state space of the process is an infinite dimensional function space.
Nonetheless, hitting questions have been studied for certain SPDE, see [DKN07,DSS10,DSS15,MT02,NV09] among others. These papers deal with the stochastic heat and wave equations either with no drift or with well behaved drift.
As for SPDE analogues of the Bessel process, the only results known to the authors are in Mueller [Mue97] and Mueller and Pardoux [MP99].
Here we assume that u(t, x) is scalar valued, and as before t > 0. But now we let x lie in the unit circle [0, 1] with endpoints identified. We also assume that u(0, x) is continuous and strictly positive. Here and throughout the paper we writeẆ (t, x) for two-parameter white noise. Suppose u satisfies the following SPDE.
The situation for vector-valued solutions u(t, x) of the stochastic heat equation is unclear. Indeed, the curve x → u(t, x) may wind around 0, and perhaps then u will contract to 0 in cases where it would ordinarily stay away from 0.
The purpose of this paper is to study hitting question for the stochastic wave equation with scalar solutions and with strong drift. As is well known, there are crucial differences between the heat and wave equations. For example, the heat equation satisfies a maximum principle while the wave equation does not. The same holds for the comparison principle, which states that if the stochastic heat equation has two solutions with the first solution initially larger than the second, then the first solution will almost surely remain larger than the second as time goes on. So while certain arguments from the heat equation case carry over, new ideas are required.
Here is the setup for our problem. Again, we let t ≥ 0, and x lies in the circle with endpoints identified. We study scalar-valued solutions u(t, x) to the following equation.
As usual, u and our two-parameter white noiseẆ depend on a random parameter ω which we suppress. As for x taking values in higherdimensional spaces, it is well known that (1.1) is well-posed only in one spatial dimensions. Indeed, in two or more spatial dimensions we would expect that the solution u only exists as a generalized function, but then it is hard to give meaning to nonlinear terms such as u −α or g(u).
Next, we define the first time that u hits 0. Let and let τ ∞ = ∞ if the set in the above definition is empty. Before stating our main theorems, we give some assumptions.

Assumptions
(i) u 0 is continuous.
(ii) There exist constants 0 < c 0 < C 0 < ∞ such that c 0 ≤ u 0 (x) ≤ C 0 for all x ∈ I. (iii) u 1 is continuous on I and hence bounded. (iv) There exist constants 0 < c g < C g < ∞ such that c g ≤ g(y) ≤ C g for all y ∈ R.
Here are our main theorems.
Theorem 1. Suppose that u(t, x) satisfies (1.1), and that the above assumptions hold. Then α > 3 implies That is, u does not hit 0.
Here is the plan for the paper. In Section 2 we give a rigorous formulation of (1.1); in particular, the solution is only defined up to the first time t that u(t, x) = 0 for some x, since u −α (t, x) blows up there. The same is true for the stochastic heat equation discussed earlier. In Section 3 we prove Theorem 1, and in Section 4 we prove Theorem 2.
Note the gap between α < 1 and α > 3. Since there is no comparison principle for the wave equation, we cannot be certain that there exists a critical value α 0 such that u can hit 0 for α < α 0 but not for α > α 0 . We strongly believe in the existence of such a critical value, but we leave the existence and identification of α 0 as an open problem.
2. Technicalities 2.1. Rigorous Formulation of the Wave SPDE. For the most part we follow Walsh [Wal86] although we could also use the formulation found in Da Prato and Zazbczyk [DPZ92].
First we recall the definition the one-dimensional wave kernel on x ∈ R.
See [Eva98] for this classical material. If we regard S(t, x) as a Schwartz distribution, then for t ≥ 0 we can write From now on, we interpret such expressions as Schwartz distributions. Now we switch to the circle x ∈ I, as defined earlier. It is also a classical result that for x ∈ I, the wave kernel S I and its time derivative are given by Again, we regard ∂ t S I as a Schwartz distribution. Let w(t, x) be the solution of the linear deterministic wave equation on x ∈ I, with the same initial data as u. That is, with periodic boundary conditions, so that where expressions such as x − y and x − t − y are interpreted using arithmetic modulo J. Using Duhamel's principle, if u −α and g(u(s, y))Ẇ were smooth, we could write If u −α had no singularities, we could use this mild form to give rigorous meaning to (1.1), where we define final double integral using Walsh's theory of martingale measures, see [Wal86]. One could also use the Hilbert space theory given in Da Prato and Zabczyk [DPZ92].
To deal with the singularity of u −α , we use truncation and then take the limit as the truncation is removed. For N = 1, 2, . . . define u N (t, x) as the solution of Note that if α > 0, then [u ∨ (1/N)] −α is a Lipschitz function of u. It is well known that SPDE such as (2.2) with Lipschitz coefficients unique strong solutions valid for all time, see [Wal86], Chapter III. It follows for each N = 1, 2, . . . that (2.2) has a unique strong solution u N valid From the definition, we see that almost surely for all t ∈ [0, τ N 1 ∧ τ N 2 ) and x ∈ I. Here a ∧ b is the minimum of a and b. It also follows that τ 1 ≤ τ 2 ≤ · · · and so we can almost surely define We allow the possibility that τ = ∞. Note that this definition of τ is consistent with the definition given in the introduction. So, for t < τ and x ∈ I, we can define since for t < τ and x ∈ I the sequence u 1 (t, x), u 2 (t, x), . . . does not vary with N after a finite number of terms. It follows that u(t, x) satisfies (2.1) for 0 ≤ t < τ . Finally, we define u(t, x) for all times t by defining Here ∆ is a cemetary state.

2.2.
Multi-parameter Girsanov Theorem. The proof of Theorem 2 is based on Girsanov's theorem for two-parameter white noise. This approach was used earlier in Mueller and Pardoux [MP99] for the stochastic heat equation, but we need to do some work to adapt the argument to the stochastic wave equation. Girsanov's theorem will allow us to remove the drift from our equation (1.1), at least up to time τ . If this Girsanov transformation gives us an absolutely continuous change of probability measure, then we only need to verify that the stochastic wave equation (1.1) without the drift has a positive probability of hitting 0. Assume that our white noiseẆ (t, x) and hence also u(t, x) is defined on a probability space (Ω, F , P). As in Walsh [Wal86], we definė W (t, x) in terms of a random set function W (A, ω) on measurable sets A ⊂ [0, ∞) × I. Let (F t ) t≥0 be the filtration defined by Nualart and Pardoux [NP94] give the following version of Girsanov's theorem.
Theorem 3. Let T > 0 be a given constant, and define the probability measure P T to be P restricted to sets in F T . Suppose that W is a space-time white noise random measure on [0, T ] × R with respect to P T , and that h(t, x) is a predictable process such that the exponential process is a martingale for t ∈ [0, T ]. Then the measure is a space-time white noise random measure on [0, T ]×R with respect to the probability measure Q T , where Q T and P T are mutually absolutely continuous and We recall Novikov's sufficient condition for E h (t) to be a martingale.
Proposition 1. Let h(t, x) be a predictable process with respect to the filtration 2.3. Hölder continuity of the stochastic convolution. For an almost surely bounded predictable process ρ(t, x), we define the stochastic convolution as follows.
Note that the double integral in (2.1) is equal to N g(u) (t, x) for t < τ . We conveniently define g(∆) = 0, so that N g(u) (t, x) is defined for all time.
The proofs of both main theorems depend on the Hölder continuity of N g(u) (t, x). Although such results are common in the SPDE literature, unfortunately we could not find the exact result we needed. So for completeness, we state it here.
Theorem 4. Let ρ(t, x) be an almost surely bounded predictable process. For any T > 0 and β < 1/2, there exists a random variable Y with finite expectation, with E|Y | depending only on β and T , such that We will prove Theorem 4 in the appendix.

Proof of Theorem 1
3.1. Outline and Preliminaries. We write the mild solution to (1.1) in the following form: We will prove Theorem 1 by contradiction. First, we assume that τ < ∞ with positive probability. Then, on the sample paths where this is the case (i.e., all u(ω) such that τ (ω) < ∞), we go backwards in time from where u hits zero. The upward drift term D u (t, x) will then push downwards, since we are going backwards in time. We show that this downward push must overwhelm the modulus of continuity of the N u (t, x) term, implying the existence of another time τ 1 < τ such that u hits zero at τ 1 . However, this contradicts the minimality of τ , thus proving the theorem.

A Regularity Lemma. Let
By assumption, P(A) > 0. We then show the following: for any β < 1/2. The Holder constant is a random variable depending only on β and ω.
Proof. Let β < 1/2 be given. Then by (2.6) we know that N u (t, x) is almost surely β-Holder continuous on [0, τ ) × [0, J], with random Holder constant Y depending only on β and τ . Since u 1 is continuous on I, the Riemann integral As the Holder constant of V u depends only on u 0 and u 1 , the Holder constant of V u + N u is a random variable depending only on β.
3.3. The Backwards Light Cone. Given (t, x) ∈ R + ×R, define the backwards light cone as Note that the light cone cannot include points (s, y) for which s > t. It follows that D u (t, x) can be rewritten as Proof. Since u(s, y) > 0 on [0, τ ), using (3.2) the result follows from the fact that L(s, y) L(t, x).
Using Lemma 1, on the event A we define Y to be a (random) 1/2 − ǫ Holder constant of V u (t, x) + N u (t, x), depending only on ǫ. By our choice of ǫ, the exponent of R in the expression is negative. Hence, on A we can pick a sufficiently small random R > 0, depending on ǫ and Y , such that both Finally, on A we pick a random δ > 0 sufficiently small such that both which is possible since since u(t, x) is continuous in t for t < τ . Here, τ δ need not be a stopping time. Note that τ δ is the first time that u(t, x) reaches δ, and that by continuity of We define the differences and for all (t, x) ∈ L(τ δ , x δ ), we decompose We recall that by construction, From Lemma 2, we find that ∆D(t, x) < 0 almost surely. Hence, for all (t, x) ∈ L (τ δ , x δ ) we obtain the bound almost surely on A. We define the sector noting from (3.4) and (3.6) that t > 0 on B R . We then denote the curved part of the boundary of B R by Then for all (t, x) ∈ ∂B r , using (3.2), (3.9), and (3.5) we find that almost surely on A. From (3.4) and (3.11) it then follows that u(t, x) < 0 for all (t, x) ∈ ∂B R , almost surely on A.
Since P(A) > 0 by assumption, the event that u(t, x) < 0 for all (t, x) ∈ ∂B R occurs with positive probability. However, since R > 0, we know that t < τ δ < τ for all (t, x) ∈ ∂B R , which is a contradiction, since τ is defined to be the first hitting time for u(t, x) ≤ 0. Hence we conclude that P(A) = 0.
This finishes the proof of theorem 1.
Here x ∈ [0, J], as before. Since there are no singular terms in (4.1), we can give this equation rigorous meaning using the mild form: where w(t, x) is as before, the solution to the deterministic wave equation.
First we verify that v(t, x) can hit 0.
Lemma 3. Suppose that v(t, x) is a solution to (4.2). Then By the almost sure continuity of v(t, x) (see [Wal86]) Chapter III, it suffices to show that x, y)dy = t by the definition of the one-dimensional wave kernel, and since J 0 Here we have used the stochastic Fubini theorem (see [Wal86], Theorem 2.6) to change the order of integration in the double integral. Let us define N v (t) as the double integral: The question would be easy if g ≡ 1, as N v (t) would be a Gaussian variable, with a positive probability of taking values below any desired level. Since this is not necessarily the case, we use another Girsanov transformation to bound N v (t) by a Gaussian process.
Fix t > 0. Choose K sufficiently large so that Using Theorem 3, we defineW as aP white noise, where P andP are equivalent and (s, y))Kdyds.
Since g(v(s, y)) is bounded below by c g > 0, we have: Hence to show (4.3), it suffices to prove that and since P andP are equivalent, we can show instead that We define the process M t (r) = Since g is bounded, M t (r) is an F r -martingale in r, for r ≤ t. Hence, from Theorem V.1.6 in Revuz and Yor [RY99], there exists a onedimensional standard Brownian motion B such that M t (r) = B (τ (r)), where the time change τ (r) is given by the predictable process: Then let so we have τ (t) ≤ L. Using this, we find that: Due to (4.4), we can use the reflection principle to find that from which (4.5) follows, and the proof of Lemma 3 is complete.
and for a constant T > 0, let We then obtain the following Girsanov transformation: Lemma 4. For each m ∈ N, the measures P Tm(u) u and P Proof. First, we note that h(v Tm(v) (t, x)) satisfies the condition given in (2.5). Then, define the probability measure Q Tm(v) by the derivative Then, from Theorem 3, it follows that where the last term is a Walsh integral with respect to the underlying measure Q Tm(v) . However, these are just the paths of u Tm(u) (t, x), so the measure Q Tm(v) = P Tm(u) u . Then Lemma 4 follows.
Note that Theorem 2 follows from the following lemma, which we prove by using the regularity of the stochastic wave equation. and observe that on the event A(K), where µ denotes Lebesgue measure. Define a constant ǫ > 0 such that 0 < 2ǫ < 1 − α and for each n ∈ N, consider the rectangle Then for each (t, x) ∈ D n , define the grids of points: Let # denote the number of points in a set, and define the strip Then we have Since v(t, x) is continuous on [0, T ] × [0, J] and inf x∈[0,J] u 0 (x) > 0, we have that µ (J n ) = 0 for sufficiently large random n. Hence, (4.13) ∞ n=0 2 2α(n+1) K −2α µ (J n ) < ∞ almost surely. We now place a bound on # (s, y) ∈ Γ n (t, x) : v(s, y) ≤ 2 −n K in the upcoming lemmas.

The Shifted Equation.
Let (t, x) be an arbitrary point in D n , as defined in (4.10), and let θ be the time shift operator, defined by θ s W (dxdt) = W (dxd(t + s)).
Then for given (s, y) ∈ Γ n (t, x), define Now, we take the approach of considering W as a cylindrical Wiener process, as described in [DPZ92]. Furthermore, by Theorem 9.15 on page 256 of Da Prato and Zabczyk [DPZ92], there is a version of our solution Φ t which is a strong Markov process with respect to the Brownian filtration (F t ) t≥0 .
Using the strong Markov property of solutions, we restart the equation at time s − n : (4.14) v(s, y) = 1 2 v s − n , y + λ n + v s − n , y − λ n + J 0 S I (λ n , y, z) ∂v ∂t s − n , z dz Here, ∂v ∂t is regarded as a Schwartz distribution. We analyze (4.14) term by term. Decompose More specifically, • First, we take V n to be the first two terms, representing the contribution to v(s, y) from the shifted initial conditions (both position and velocity). • Next, we realize the stochastic term as the sum of a conditionally Gaussian term and an error term. The former is the stochastic term integrated over the light cone contained in the square {(s, y) + D n }, with the diffusion coefficient g frozen at v(s − n , y). We call this term the noise term, N n . • Finally, as mentioned above the error term E n is the difference between the stochastic term of v(s, y) minus the noise term defined above.
As alluded to above, the noise term can be rewritten as: where c 2 n is the quadratic variation of the above double integral and Z is a standard normal random variable. Moreover, for sufficiently small λ n relative to J, we have: Then we assert the following: To prove this lemma, we first establish a bound on the error term E n . Recall its definition: Note that in the integrals above, S I (λ n −r, y, z) = 0 outside of the light cone |z − y| ≤ λ n . Thus, we restrict the domain of integration of z: We define the rectangle ∆ n (s, y) = {r ∈ R + , z ∈ [0, J] : |r − s| ≤ λ n , |z − y| ≤ λ n } and let p be a positive integer. Then it follows that and since the integrand above is continuous in λ n , we can use the Burkholder-Davis-Gundy inequality to obtain: Since g is Lipschitz and since (s − n , y) ∈ ∆ n (s, y), With this bound, we get: Recall that v(s, y) is almost surely β-Holder continuous for any β < Recalling (4.11), we then bound the integral: Proof of Lemma 6. Recalling that # {Γ n (t, x)} λ −2 n = 2 (2−4ǫ)n−1 , we find: P |E n (s, y)| > 2 −n 2 (2−4ǫ)n P |E n (s, y)| > 2 −n By Markov's inequality, we can continue as follows, n after which we use (4.20) to obtain the bound: With a similar decomposition, we obtain: P (C c n ) 2 (4−4ǫ)n P |N n (s, y)| > 2 −(1−3ǫ)n Recalling (4.15), this implies: P (C c n ) 2 (4−4ǫ)n P g v s − n , y Z > 2 (1−2ǫ)n 2 −(1−3ǫ)n 2 (4−4ǫ)n P |Z| > C −1 g 2 ǫn where Z is a standard normal random variable. Now, we use a standard tail estimate for the normal (often called the Chernoff bound) to conclude P |Z| > C −1 g 2 ǫn ≤ 2 exp −C −2 g 2 2ǫn−1 and it follows that ∞ n=1 P (C c n ) converges.

A Counting Lemma.
Lemma 7. For each K > 0 there exists a constant c K such that for every n ∈ N and (t, ) ∈ D n , The proof of Lemma 7 will require several preliminary steps. We fix (t, x) and order the points in Γ n (t, x) lexicographically, calling the ith point (s i , y i ) for some i ∈ I(t, x) = {1, 2, . . . , #{Γ n (t, x)}} -i.e., if i < j then s i ≤ s j and if s i = s j , then x ≤ x i < x j mod J. For given (t, x), we define the set ∆ n (s, y) as follows: where the interval [x − λ n , y − λ n ] on S is taken modulo J, wrapping around whenever x − λ n > y − λ n . (Note that this is not the same as the previously defined ∆ n (s, y)). Let F n i be the σ-algebra generated byẆ in the set ∆ n (s i , y i ). Then V n (s i , y i ) is F n i -measurable for all i ∈ N. Recall from (4.15) that (4.21) N n (s i , y i ) = g v (s i ) − n , y i c n Z i where c n = 2 −(1−2ǫ)n−1 and Z i ∼ N (0, 1) is F n i+1 -measurable but independent of F n i . Let P n i denote the conditional probability with respect to F n i and let δ > 1 be a constant depending only on K. Define v n (s, y) = V n (s, y) + N n (s, y).
We now prove the following lemma: Lemma 8. There exists d K > 0 such that for all i ∈ I(t, x), Proof. From the definition of conditional probability, the left hand side of (4.22) is: Using (4.21), we find that y i ) and using (4.16), we find that .
Since the denominator in (4.23) is less than or equal to 1, we can bound H below by its numerator: Using the decomposition and the assumption in (4.24), we find Since ρ n,i ≤ 2C g for all n, we note that for all K > 0, ρ n,i 2 −2ǫn (K +2) → 0 as n → ∞. So for sufficiently large n (depending on K), H ≥ 1/3. Hence in the case given by (4.25), Lemma 8 follows.
Proof of Lemma 7. Define From the definitions of A n and Γ, it follows that ξ n is bounded by: |N n (s, y)| ≤ 2 −2(1−3ǫ)n .
Out of the set of points on Γ n such that v n ≤ 2 −n (K + 1), we want to find the points where v n < −2 −n , which would force v to be negative. Thus we define the event D k = v n s σn(k) , y σn(k) ≤ −2 −n and for k ∈ N, we define the indicator random variable From Lemma 8, it is clear that and moreover, since V i and N i are F n j -measurable for all i < j, we can also use Lemma 8 to find that P I k = 1 I 1 , . . . , I k−1 ≥ d K for k > 1. Finally, let Since v(s, y) ≥ 0 for all (s, y) ∈ Γ n , it follows that ξ n ≤ Eσ n .
Note that the I k 's are not independent. We couple {I k } with a sequence of independent random variables {Y k } as follows: Let {U k } k≥1 be a sequence of mutually independent random variables that are globally independent of the I k 's and have uniform law on [0, 1]. Then define Y k =    0 if I k = 0 or U k > d K /P I k = 1 I 1 , . . . , I k−1 1 if I k = 1 and U k ≤ d K /P I k = 1 I 1 , . . . , I k−1 for k ≥ 1. Then clearly, Y k ≤ I k and for k > 1, and since Y k ≤ I k , it follows that σ n ≤σ. So 4.4. Lemma 5, Conclusion. Finally, we cite a measure-theoretic result related to the Borel-Cantelli Lemma: Lemma 9. Let {X n } be a sequence of R ≥0 -valued random variables, and {F n } be a sequence of events, such that both: Proof. Let F = { ∞ n=0 X n = +∞}. Then on the event F ∩ (lim inf F n ), we have ∞ n=0 X n 1 Fn = +∞. So from the second condition, we get P(F ∩ lim inf F n ) = 0. However, from the first condition and Borel-Cantelli, we find: Implying that P(F) = 0, which is our desired result.
A.0.1. Space Difference. To bound N(t, x + k) − N(t, x) we first write and note that for all r, the following stochastic integral is a martingale over t: We fix p > 2 and use Burkholder's inequality to find a constant C p such that |ρ(s, y)| 2 |S I (r − s, x + k, y) − S I (r − s, x, y)| 2 dyds p/2 for all r. As C p does not depend on r, we can set r = t to obtain . Now we use Holder's Inequality with exponents p 2 and p p−2 : Since ρ is almost surely bounded, the expectation E t 0 S g (v(s, y)) p dyds is bounded by a constant depending on T and p. So we obtain: As usual, the notation a(x) p,T b(x) means that a(x) ≤ C p,T b(x).
We use the inequality For the second term, we start with Holder's inequality again: Using the inequality N n=1 a n p p,N N n=1 a p n one last time, we continue the inequality: Finally, we recall Kolmogorov's continuity theorem for multiparameter processes [Wal86], Corollary 1.2.
Theorem 5 (Kolmogorov). Let R be a rectangle in R n and {X t , t ∈ R} be a real-valued stochastic process. Suppose there exist a, b, K all positive such that for all s, t ∈ R E |X t − X s | a ≤ K|t − s| n+b .

Then,
(1) X has a continuous realization; (2) there exist a constant C depending only on a, b, n and a random variable Y such that with probability one, Setting a = p and b = p/2 − 3, we obtain ≤ Y h 1/2−3/p + k 1/2−3/p where E[Y ] < ∞ and depends only on p and T . Then since β < 1/2, we can set p = 6 1−2β and the conclusion follows.