EElectron. Commun. Probab. (2012), no. 0, 1–2.DOI: vVOL-PID ISSN:
ELECTRONICCOMMUNICATIONSin PROBABILITY
Exit Boundaries of Multidimensional SDEs
Russell Lyons * Abstract
We show that solutions to multidimensional SDEs with Lipschitz coefficients anddriven by Brownian motion never reach the set where all coefficients vanish unlessthe initial position belongs to that set.
Keywords:
Inaccessible; Lipschitz; singular.
AMS MSC 2010:
The classification of isolated singular points of a 1-dimensional SDE driven by Brown-ian motion is complete and exhibits several types of behavior: see [1, Fig. 2.2] for a goodsummary. For example, as has long been known, if X is a (weak) solution to E x ( σ, with σ − being nonzero and locally integrable in some interval (0 , a ] and x ∈ (0 , a ) , then theprobability that X t ever reaches 0 is positive (i.e., 0 is accessible ) iff (cid:82) a y σ ( y ) − d y < ∞ .Much less is known in higher dimensions. In particular, the following theorem that makesthe usual assumption of Lipschitz coefficients seems to be new: Theorem.
Let d, m ∈ N + . Let B = ( B (1) , . . . , B ( m ) ) be m -dimensional Brownian motion.Let σ : R d → M d × m ( R ) and b : R d → R d be Lipschitz. Write Λ := { x ∈ R d ; σ ( x ) = 0 , b ( x ) = 0 } . Suppose that X solves E x ( σ, b ) , i.e., X t = x + (cid:90) t σ ( X s ) d B s + (cid:90) t b ( X s ) d s ( t ≥ . If x / ∈ Λ , then P [ ∀ t ≥ X t / ∈ Λ] = 1 . In other words, the set Λ is inaccessible. Proof.
We use the Frobenius norm (cid:107) M (cid:107) := (cid:112) Tr( M ∗ M ) for a matrix, M . For A > ,define the stopping time T A := inf (cid:8) t ≥ (cid:107) σ ( X t ) (cid:107) + (cid:107) b ( X t ) (cid:107) = A (cid:9) . Fix
A > . For k ∈ N + , write S k := T A/ k +1 ∧ T A/ k − . * Department of Mathematics, 831 E. 3rd St., Indiana University, Bloomington, IN 47405-7106. E-mail: [email protected] . Partially supported by NSF grant DMS-1612363. a r X i v : . [ m a t h . P R ] F e b xit Boundaries of Multidimensional SDEsIf x is such that (cid:107) σ ( x ) (cid:107) + (cid:107) b ( x ) (cid:107) = A/ k , then ∀ t ≥ E x (cid:2) (cid:107) x − X t ∧ S k (cid:107) [ S k ≤ (cid:3) ≤ ( m + 1) E x (cid:20) d (cid:88) i =1 m (cid:88) j =1 (cid:16)(cid:90) t ∧ S k σ ( X u ) i,j d B ( j ) u (cid:17) + d (cid:88) i =1 (cid:16)(cid:90) t ∧ S k b ( X u ) i d u (cid:17) [ S k ≤ (cid:21) ≤ ( m + 1) E x (cid:20)(cid:90) t ∧ S k (cid:107) σ ( X u ) (cid:107) d u (cid:21) + ( m + 1) E x (cid:20)(cid:90) t ∧ S k (cid:107) b ( X u ) (cid:107) d u (cid:21) ≤ ( m + 1) · A k − · t and E x (cid:104)(cid:12)(cid:12) (cid:107) σ ( x ) (cid:107) + (cid:107) b ( x ) (cid:107) − (cid:107) σ ( X t ∧ S k ) (cid:107) − (cid:107) b ( X t ∧ S k ) (cid:107) (cid:12)(cid:12) ; S k ≤ (cid:105) ≤ E x (cid:104) (cid:107) σ ( x ) + σ ( X t ∧ S k ) (cid:107) · (cid:107) σ ( x ) − σ ( X t ∧ S k ) (cid:107) + (cid:107) b ( x ) + b ( X t ∧ S k ) (cid:107) · (cid:107) b ( x ) − b ( X t ∧ S k ) (cid:107) ; S k ≤ (cid:105) ≤ E x (cid:104)(cid:0) (cid:107) σ ( x ) (cid:107) + (cid:107) σ ( X t ∧ S k ) (cid:107) + (cid:107) b ( x ) (cid:107) + (cid:107) b ( X t ∧ S k ) (cid:107) (cid:1) · K · (cid:107) x − X t ∧ S k (cid:107) ; S k ≤ (cid:105) ≤ · (cid:16) A + 2 A k (cid:17) / · K · E x (cid:2) (cid:107) x − X t ∧ S k (cid:107) ; S k ≤ (cid:3) / , where K is a bound for the Lipschitz constants. If, in addition, t ≤ and S k ≤ t , then (cid:12)(cid:12) (cid:107) σ ( x ) (cid:107) + (cid:107) b ( x ) (cid:107) − (cid:107) σ ( X t ∧ S k ) (cid:107) − (cid:107) b ( X t ∧ S k ) (cid:107) (cid:12)(cid:12) [ S k ≤ ≥ A k +1 . Putting these inequalities together, we obtain ∀ t ≤ P x [ S k ≤ t ] ≤ k +1 A · (cid:16) A k (cid:17) / · K · (cid:114) ( m + 1) · A k − · t = C √ t for some constant, C , depending only on m and K .Choose t ∈ (0 , so that C √ t ≤ / . Then by the strong Markov property, if k ∈ N + , A > , and x ∈ R d , (cid:107) σ ( x ) (cid:107) + (cid:107) b ( x ) (cid:107) ≥ A/ k = ⇒ P x [ T A/ k +1 ≥ t ] ≥ / . Given x / ∈ Λ , choose A := (cid:107) σ ( x ) (cid:107) + (cid:107) b ( x ) (cid:107) and express the time to reach Λ as (cid:80) k ≥ (cid:0) T A/ k +1 − T A/ k (cid:1) . By the strong Markov property, infinitely many of these termsare at least t a.s., whence the total time is infinite a.s. Acknowledgment.
We are grateful to Jean-François Le Gall for showing us a similaridea in a different context.
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