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Vol. 7 (2002) Paper no. 19, pages 1–19.
Journal URL
http://www.math.washington.edu/~ejpecp/
Paper URL
http://www.math.washington.edu/~ejpecp/EjpVol7/paper19.abs.html
TRANSIENCE AND NON-EXPLOSION OF CERTAIN
STOCHASTIC NEWTONIAN SYSTEMS
V.N. Kolokol’tsov
Department of Computing and Mathematics
Nottingham Trent University, Burton Street, Nottingham
NG1 4BU, UK
vk@yquem.ntu.ac.uk
R.L. Schilling and A.E. Tyukov
School of Mathematical Sciences
University of Sussex, Falmer, Brighton
BN1 9QH, UK
R.Schilling@sussex.ac.uk and A.Tyukov@sussex.ac.uk
Abstract: We give sufficient conditions for non-explosion and transience of the solution (x t , pt )
(in dimensions > 3) to a stochastic Newtonian system of the form
(
dxt = pt dt
(xt )
dt −
dpt = − ∂V∂x
∂c(xt )
∂x
dξt
,
where {ξt }t>0 is a d-dimensional Lévy process, dξt is an Itô differential and c ∈ C 2 (Rd , Rd ),
V ∈ C 2 (Rd , R) such that V > 0.
Keywords and phrases: Stochastic Newtonian systems; Lévy processes; α-stable Lévy processes; Transience; Non-explosion
AMS subject classification (2000): 60H10, 60G51, 60J75, 70H99, 60J35
Submitted to EJP on June 14, 2002. Final version accepted on October 2, 2002.
1
Introduction
This work contributes to the series of papers [13, 15], [3, 4], [6], [20] and [19] which are devoted
to the qualitative study of the Newton equations driven by random noise. For related results
see also [5], [23], [26, 27], [1], [22] and the references given there. Newton equations of this type
are interesting in their own right: as models for the dynamics of particles moving in random
media (cf. [25]), in the theory of interacting particles (cf. [28], [29]) or in the theory of random
matrices (cf. [24]), to mention but a few. On the other hand, the study of these equations
fits nicely into the the larger context of (stochastic) partial differential equations, in particular
Hamilton-Jacobi, heat and Schrödinger equations, driven by random noise (see [32, 33] and
[14, 16, 17, 18]).
In most papers on this subject the driving stochastic process is a diffusion process with continuous sample paths, usually a standard Wiener process. Motivated by the recent growth of
interest in Lévy processes, which can be observed both in mathematics literature and in applications, the present authors started in [20] and [19] the analysis of Newton systems driven by
jump processes, in particular symmetric stable Lévy processes. In [20] we studied the rate of
escape of a “free” particle driven by a stable Lévy process and its applications to the scattering
theory of a system describing a particle driven by a stable noise and a (deterministic) external
force.
In this paper we study non-explosion and transience of Newton systems of the form


dxt = pt dt
,
∂V (xt )
∂c(xt )

dpt = −
dt −
dξt
∂x
∂x
(1)
1
where
ξtd ) is a d-dimensional Lévy process, c ∈ C 2 (Rd , Rd ), V ∈ C 2 (Rd ), V > 0
³ ξt = (ξt´, . . . , P
t)
i (xt )
:= dj=1 ∂c∂x
dξtj is an Itô stochastic differential.
and ∂c(x
∂x dξt
j
i
In Section 3 we give conditions under which the solutions do not explode in finite time. For
(α)
symmetric α-stable driving processes ξt = ξt we show in Section 4 that the solution process
of the system (1) is always transient in dimensions d > 3. We consider it as an interesting open
problem to find necessary and sufficient conditions for transience and recurrence for the system
(1) in dimensions d < 3. Even in the case of a driving Wiener process (white noise) only some
partial results are available for d = 1, see [4, 3].
2
Lévy Processes
The driving processes for our Newtonian system will be Lévy processes. Recall that a ddimensional Lévy process {ξt }t>0 is a stochastic process with state space Rd and independent
and stationary increments; its paths t 7→ ξt are continuous in probability which amounts to
saying that there are almost surely no fixed discontinuities. We can (and will) always choose
a modification with càdlàg (i.e., right-continuous with finite left limits) paths and ξ 0 = 0. Unless otherwise stated, we will always consider the augmented natural filtration of {ξ t }t>0 which
satisfies the “usual conditions”. Because of the independent increment property the Fourier
2
transform of the distribution of ξt is of the form
E(eiηξt ) = e−tψ(η) ,
t > 0, η ∈ Rd ,
with the characteristic exponent ψ : Rd → C which is given by the Lévy-Khinchine formula
Z
¡
¢
ψ(η) = −iβη + ηQη +
1 − eiyη + i yη 1{|y|<1} ν(dy).
(2)
Rd \{0}
Here β ∈ Rd , Q = (qij ) ∈ Rd×d is aR positive semidefinite matrix and ν is a Lévy measure, i.e., a
Radon measure on Rd \ {0} with y6=0 |y|2 ∧ 1 ν(dy) < ∞. The Lévy-triple (β, Q, ν) can also be
used to obtain the Lévy decomposition of ξt ,
ZZ
ZZ
Q
ξt = W t +
y Ñ (dy, ds) +
y N (dy, ds) + βt
(3)
[0,t]×{0<|y|<1}
[0,t]×{|y|>1}
P
where ∆ξt := ξt − ξt− , ξ0− := ξ0 , N (dy, ds) =
06t6s 1{∆ξt 6=0} δ(∆ξt ,t) (dy, ds), is the canonical jump measure, Ñ (dy, ds) = N (dy, ds) − ν(dy) ds is the compensated jump measure,
WtQ is¡ a Brownian
motion with
¢ covariance matrix Q and βt is a deterministic drift with
P
β = E ξ1 − s61 ∆ξs 1{|∆ξs |>1} . Notice that the first two terms in the above decomposition
(3) are martingales.
Lemma 1. Let {ξt }t>0 be a d-dimensional Lévy process whose jumps are bounded by 2R. Then
Z
E([ξ i , ξ j ]t ) 6 t max |qij | + t
|y|2 ν(dy),
t > 0,
16i,j6d
0<|y|<2R
where [ξ i , ξ j ]• denotes the quadratic (co)variation process.
This Lemma is a simple consequence of the well-known formula
¶
µ
µ
Z
X
¡ i j ¢
i
j
i
j
E [ξ , ξ ]t = E [W , W ]t +
∆ξs ∆ξs = t qij +
s6t
i j
¶
y y ν(dy) .
|y|<2R
It is well-known that Lévy processes are Feller processes. The infinitesimal generator (A, D(A))
of the process
(more precisely: of the associated Feller semigroup) is a pseudo-differential oper¯
ator A¯C ∞ (Rd ) = −ψ(D) with symbol −ψ, i.e.,
c
−ψ(D)u(x) := −(2π)−d/2
Z
ψ(η)b
u(η)eiyη dη,
u ∈ Cc∞ (Rd ),
(4)
Rd
where u
b(η) denotes the Fourier transform of u. The test functions C c∞ (Rd ) are an operator core.
Later on, we will also use the following simple fact.
3
x
), R > 1. Then
Lemma 2. Let u ∈ Cc∞ (Rd ) and uR (x) := Ru( R
Z
u(η)| dη = Cψ,u R
|ψ(D)uR (x)| 6 Cψ R (1 + |η|2 ) |b
Rd
uniformly for all x ∈ Rd with an absolute constant Cψ,u .
Proof. Observe that u
bR (η) = Rd+1 u
b(Rη). Therefore,
¯
¯
¯
¯Z
¯
¯
−d/2 ¯
ixη
bR (η) dη ¯¯
|ψ(D)uR (x)| = (2π)
¯ e ψ(η) u
¯ d
¯
R
Z
6 (2π)−d/2 R Rd |ψ(η) u
b(Rη)| dη
Rd
= (2π)
−d/2
R
Z
Rd
6 (2π)
−d/2
¯
¯ ¡η¢
¯ dη
¯ψ
u
b
(η)
R
Cψ R
Z ³
Rd
6 (2π)−d/2 Cψ R
Z ³
Rd
¯ ¯2 ´
1 + ¯ Rη ¯ |b
u(η)| dη
´
1 + |η|2 |b
u(η)| dη,
where we used that |ψ(η)| 6 Cψ (1 + |η|2 ) for all η ∈ Rd with some absolute constant Cψ > 0.
b is a rapidly decreasing function which means that the integral in the last
Since u ∈ Cc∞ (Rd ), u
line is finite.
Our standard references for the analytic theory of Lévy and Feller processes is the book [10]
by Jacob, see also [11]; for stochastic calculus of semimartingales and stochastic differential
equations we use Protter [30].
3
Non-explosion
Let (Xt , Pt ) = (X(t, x0 , p0 ), P (t, x0 , p0 )) be a solution of the system (1) with initial condition
(x0 , p0 ) ∈ R2d at t = 0, where ξt = (ξt1 , . . . , ξtd ) is a d-dimensional Lévy process, d > 1,
c ∈ C 2 (Rd , Rd ), V ∈ C 2 (Rd ), V > 0 and ∂c/∂x is uniformly bounded. Clearly, these conditions
ensure local (i.e., for small times) existence and uniqueness of the solution, see e.g., [30].
The random times
Tm := inf{s > 0 : |Xs | ∨ |Ps | > m}
(5)
are stopping times w.r.t. the (augmented) natural filtration of the Lévy process {ξ t }t>0 and so
is the explosion time T∞ := supm Tm of the system (1).
Theorem 3. Under the assumptions stated above, the explosion time T∞ of the system (1) is
almost surely infinite, i.e., P(T∞ = ∞) = 1.
4
Proof. Step 1. Let τm := inf{s > 0 : |Ps | > m} and τ∞ := supm τm . It is clear that Tm 6 τm
and so T∞ 6 τ∞ . Suppose that T∞ (ω) < t < τm (ω) 6 τ∞ (ω) for some t > 0 and m ∈ N. From
the first equation in (1) we deduce that for every k ∈ N
sup
|Xs (ω)| 6 |x0 | + t sup |Ps (ω)| 6 |x0 | + tm.
s∈[0,Tk (ω)]
s∈[0,t]
On the other hand, since Tk (ω) < T∞ (ω) < t < τ∞ (ω), we find that supk∈N sups∈[0,Tk ] |Xs (ω)| =
∞. This, however, leads to a contradiction, and so τ∞ = T∞ .
Step 2. We will show that P(τ∞ = ∞) = 1. Set H(x, p) := 12 p2 + V (x) and Ht = H(Xt , Pt ).
Since H(x, p) is twice continuously differentiable, we can use Itô’s formula (for jump processes
and in the slightly unusual form of Protter [30, p. 71, (***)]). For this observe that only the
quadratic variation of the Lévy process [ξ, ξ] := ([ξ i , ξ j ])ij ∈ Rd×d contributes to the quadratic
variation of {(Xt , Pt )}t>0 :
!
µ
¶ Ã
0
0
0 £
0
¤ =
¡ ∂c ¢
¡ ∂c ¢T ∈ R2d×2d .
[(X, P ), (X, P )] =
∂c
∂c
0 ∂x
ξ, ∂x
ξ
0 ∂x
[ξ, ξ] ∂x
Therefore,
µ
µ
¶ ¶
∂c(Xt− )
1
∂c(Xt− ) T
∂V (Xt )
dHt = Pt− dPt + tr
d[ξ, ξ]t
Pt dt + Σt ,
+
2
∂x
∂x
∂x
where
Σt =
¢
1 X ¡ 2
2
Ps − Ps−
− 2Ps− (Ps − Ps− ) − (Ps − Ps− )2 = 0.
2
06s6t
The first equation in (1), dXt = Pt dt, implies that Xt is a continuous function; the second
equation, dPt = −∂V (Xt )/∂x dt − ∂c(Xt )/∂x dξt , gives
µ
µ
¶ ¶
∂c(Xt )
1
∂c(Xt ) T
∂c(Xt )
dHt = −Pt−
dξt + tr
d[ξ, ξ]t
.
(6)
∂x
2
∂x
∂x
Let σR := inf{t > 0 : |ξt | > R} be the first exit time of the process {ξt }t>0 from the ball BR (0).
Then
σ = σ`,m,R := ` ∧ σR ∧ τm ,
`, m ∈ N,
is again a stopping time and we calculate from (6) that
¶ ¶
µ
Zσ−
Zσ− µ
∂c(Xt )
∂c(Xt )
∂c(Xt ) T
1
tr
Hσ− − H0 = − Pt−
dξt +
d[ξ, ξ]t
∂x
2
∂x
∂x
0
(7)
0
= I + II.
Step 3. Recall that −ψ(D) is the generator of the Lévy process ξt . We want to estimate
|E(I)|. For this purpose choose a function φ¡ ∈¢Cc∞ (Rd , Rd ) such that φ(x) = x if |x| 6 1,
x
. Clearly,
supp φ ⊂ {x : |x| 6 2} and define φR (x) = Rφ R
φR (ξt ) = ξt ,
5
t < σR ,
(8)
and, since φR ∈ Cc∞ (Rd ) ⊂ D(A) is in the domain of the generator of ξt , we find that
MtφR
Zt
:= φR (ξt ) +
ψ(D)φR (ξs )ds
(9)
0
φR
is an L2 -martingale (w.r.t. the natural filtration of {ξt }t>0 ). The stopped process (Mt∧τ
)
m ∧` t>0
2
is again an L -martingale for fixed m, ` ∈ N. We can now use (8) and (9) to get
Zσ−
Zσ−
∂c(Xt )
∂c(Xt )
φR
I = − Pt−
dMt∧τm ∧` + Pt−
ψ(D)φR (ξt ) dt = I0 + I00 .
∂x
∂x
0
Clearly,
R•
0
0
φR
Pt− (∂c(Xt )/∂x) dMt∧τ
is a local martingale. Since
m ∧`
· Z•
∂c(Xs )
φR
Ps−
dMs∧τ
,
m ∧`
∂x
0
Z•
∂c(Xs )
φR
dMs∧τ
Ps−
m ∧`
∂x
=
Zt
0
=
2
Ps−
µ
∂c(Xs )
∂x
0
t∧τ
Zm ∧`
2
Ps−
0
µ
¶2
¸
t
d[M•φR , M•φR ]s∧τm ∧`
∂c(Xs )
∂x
¶2
d[M•φR , M•φR ]s∧τm ∧`
we find for every t > 0
¯ · Z•
¸
Z•
¯
∂c(Xs )
∂c(X
)
s
φ
φR
R
¯E
,
P
dM
dM
P
s−
s−
s∧τm ∧`
s∧τm ∧`
¯
∂x
∂x
0
0
t
¯
¯
¯
¯
° °2
° °
2 ° ∂c °
6 m ° ° E [M•φR , M•φR ]t < ∞,
∂x ∞
where we used that |Ps− | 6 m if s 6 ` ∧ τm and that MtφR is an L2 -martingale. This shows that
R•
φR
is a martingale (cf. [30], p.66 Corollary 3) and we may apply optional
0 Pt− (∂c(Xt )/∂x) dMt
stopping to the bounded stopping time σ to get
µ Zσ
∂c(Xt )
Pt−
E(I ) = −E
dMtφR
∂x
0
µ
¶
∂c(Xσ )
φR
= E Pσ−
∆Mσ
.
∂x
0
¶
µ
∂c(Xσ )
∆MσφR
+ E Pσ−
∂x
¶
Therefore
° °
° °
¯
¯
° °
° ∂c °
¯ 0 ¯
¯
¯
2
φ
2
R
¯E(I )¯ 6 md ° ° E ¯∆Mσ ¯ 6 2mRd ° ∂c ° kφk ,
∞
° ∂x °
° ∂x °
∞
∞
6
(10)
where we used
¯
¯
¯
¯
¯∆MσφR ¯ = |φR (ξσ ) − φR (ξσ− )| 6 2Rkφk∞
and the notation
° °
° ∂c °
° °
° ∂x °
:=
∞
¯
¯
¯ ∂ci (x) ¯
¯.
¯
sup ¯
∂xj ¯
d
max
i,j=1,... ,d x∈R
Step 4. For the estimate of E(I00 ), we use Lemma 2 with u = φ to get kψ(D)φR k∞ 6 Cψ,φ , and
also σ 6 `, so
¯
¯¶
° °
µ
¯
¯ 00 ¯
¯
° °
∂c(X
)
t
¯E(I )¯ 6 Cψ,φ R E sup ¯Pt−
¯ ` 6 C2 ° ∂c ° Rm `.
(11)
¯
¯
° ∂x °
∂x
t<σ
∞
Put together, the estimates (10), (11) give
|E(I)| 6 C3 Rm `.
(12)
Step 5. We proceed with |E(II)|. From
A, B ∈ Rd×d ,
kABk∞ 6 dkAk∞ kBk∞ ,
where kAk∞ = maxi,j=1,... ,d |Aij |, we get
° °2
¶ #
µ
Zt "
° °
∂c(Xs ) T
∂c(Xs )
3 ° ∂c °
d[ξ, ξ]s
6 d ° ° k[ξ, ξ]t k∞ .
tr
∂x
∂x
∂x ∞
0
Since we have sups6t |ξs | 6 R for t < σR , the jumps |∆ξs |, s 6 t, cannot exceed 2R. Lemma 1
then shows
Z
³
´
E [ξ i , ξ j ]`∧σR − 6 `
|y|2 ν(dy) + ` kQk∞
0<|y|62R
and so
|E(II)| 6 C4 `
µ
Z
¶
|y| ν(dy) + kQk∞ .
2
0<|y|62R
(13)
Step 6. Combining (7), (12), (13) we obtain
E (Hσ− ) 6 H0 + C3 Rm ` + C4 `
µ
Z
0<|y|62R
¶
|y|2 ν(dy) + kQk∞ .
On the other hand, by Jensen’s inequality,
E (Hσ− ) =
1
1
2
2
E (Pσ−
) + E(V (Xσ− )) > E (Pσ−
)
2
2
¤2
1£
E (|Pσ− |)
>
2
¢ ¤2
1£ ¡
>
E |P`∧τm ∧σR − | 1{τm <`∧σR }
2
¢ ¤2
1£ ¡
E |Pτm − ∆Pτm | 1{τm <`∧σR }
.
=
2
7
(14)
Clearly, |Pτm | > m and, since on {s < σR } the driving Lévy process has jumps of size |∆ξs | 6 2R,
we find from (1) that
° °
° ∂c °
°
|∆Pτm | 1{τm <`∧σR } 6 2R °
° ∂x ° 1{τm <`∧σR } .
∞
Choosing m sufficiently large, say m > 2Rk(∂c/∂x)k∞ , we arrive at
¤2
1£
E (m − |∆Pτm |) 1{τm <`∧σR }
2
° ° ´
2© ¡
° ∂c °
¢ª2
1³
°
m − 2R °
>
P τm < ` ∧ σ R
.
°
°
2
∂x
E (Hσ− ) >
(15)
∞
We can now combine (14) and (15) to find
© ¡
¢ª2
P τm < ` ∧ σ R
6
2(H0 + C3 Rm`)
(m − 2Rk(∂c/∂x)k∞ )2
¡
2C4 `
+
(m − 2Rk(∂c/∂x)k∞ )2
Z
0<|y|62R
¢
|y|2 ν(dy) + kQk∞ .
Letting first m → ∞ and then R → ∞ shows P(τ∞ 6 `) = 0 for all ` ∈ N, so P(τ∞ = ∞) = 1,
and the claim follows.
4
Transience
We will now prove that the solution {(Xt , Pt )}t>0 of the Newton system (1) is transient, at
(α)
least if the driving noise is a symmetric stable Lévy process ξt = ξt with index α ∈ (0, 2).
Symmetric α-stable Lévy processes have no drift, no Brownian part and their Lévy measures
are ν(dy) = cα |y|−d−α dy, where
¡
¢
α 2α−1 Γ α+d
2
¢.
cα = α/2 ¡
(16)
π Γ 1 − α2
We restrict ourselves to presenting this particular case, but it is clear that, with minor alterations,
the proof of Theorem 6 below remains valid for any driving Lévy process with rotationally
symmetric Lévy measure.
Our proof is be based on the following result which extends a well-known transience criterion
for diffusion processes to jump processes, see for instance [8] or [21].
Denote by {Tt }t>0 the operator semigroup associated with a stochastic process and let (A, D(A))
be its generator. The full generator is the set
½
¾
Z t
b
A := (f, g) ∈ Cb × Cb : Tt f − f =
Ts g ds ,
0
b for all u ∈ D(A).
see Ethier, Kurtz [7] p. 24. It is clear that (u, Au) ∈ A
8
Lemma 4. Let {ηt }t>0 be an Rn -valued, càdlàg strong Markov process with generator (A, D(A))
b Let D ⊂ Rn be a bounded Borel set and assume that there exists a sequence
and full generator A.
n
{uk }k∈N ⊂ Cb (R ) and some function u ∈ C(Rn ), such that the following conditions are satisfied:
b and
(i) A has an extension à such that Ãuk is pointwise defined, (uk , Ãuk ) ∈ A
limk→∞ (uk , Ãuk ) = (u, Ãu) exists locally uniformly.
(ii) u > 0 and inf u > a > 0 for some a > 0.
D
(iii) u(y0 ) < a for some y0 6∈ D.
(iv) Ãu 6 0 in D c .
Then {ηt }t>0 is transient.
b we know that
Proof. Since (uk , Ãuk ) ∈ A,
Mtk
= uk (ηt ) −
Zt
k ∈ N,
Ãuk (ηs ) ds,
0
are martingales, see Ethier, Kurtz [7, p. 162, Prop. 4.1.7]. We set
τD = inf{t > 0 : ηt ∈ D} and
σR = inf{t > 0 : |ηt − η0 | > R}
and from an optional stopping argument we find for any fixed T > 0
³
´
Ey0 MτkD ∧σR ∧T = Ey0 (M0k ) = Ey0 (uk (η0 )).
On the other hand,
³
Ey0 MτkD ∧σR ∧T
´

τD ∧σ
Z R ∧T
= Ey0 uk (ητD ∧σR ∧T ) −
0

Ãuk (ηs )ds ,
and because of assumption (i) we can pass to the limit k → ∞ to get
a > u(y0 ) = lim uk (y0 )
k→∞

= lim Ey0 uk (ητD ∧σR ∧T ) −
k→∞

= Ey0 u(ητD ∧σR ∧T ) −
>E
y0
¡
¢
τD ∧σ
Z R ∧T
τD ∧σ
Z R ∧T
0
u(ητD ∧σR ∧T )
¡
¢
> E u(ητD ∧σR ∧T ) 1{τD <∞} ,
y0
¯
where we used in the penultimate step that Ãu¯Dc 6 0.
9
0

Ãuk (ηs )ds

Ãu(ηs )ds
As u ∈ C + (Rn ), we may use dominated convergence and let T → ∞ and Fatou’s Lemma to let
R → ∞. Thus,
¢
¡
¢
¡
a > u(y0 ) > lim inf Ey0 u(ητD ∧σR )1{τD <∞} > Ey0 u(ητD )1{τD <∞}
R→∞
> (inf u) Py0 (τD < ∞) > a Py0 (τD < ∞).
D
Therefore, Py0 (τD < ∞) < 1, and, see e.g [2], {ηt }t>0 is transient.
We will now turn to the task to determine the infinitesimal generator of the solution process
{(Xt , Pt )}t>0 . The following result is, in various settings, common knowledge. We could not
find a precise reference in our situation, though. Since we need some technical details of the
proof, we include the standard argument.
Lemma 5. Let {ξt }t>0 be a d-dimensional Lévy process with characteristic exponent ψ and
Lévy triple (α, Q, ν). The (pointwise) infinitesimal generator of the process (X t , Pt ) =
(X(t, x0 , p0 ), P (t, x0 , p0 )) solving (1) is of the form
µ
¶
∂u(x, p)
∂u(x, p) ∂V (x) ∂c(x)
Au(x, p) =
p−
+
β
∂x
∂p
∂x
∂x
Ã
µ
¶ µ
¶ !
∂ 2 u(x, p) ∂c(x)
∂c(x) T
1
Q
+ tr
2
∂p2
∂x
∂x
¶
Z µ
∂u(x, p) ∂c(x)
∂c(x)
+
u(x, p − ∂x y) − u(x, p) +
y 1{|y|<1} ν(dy).
∂p
∂x
Rd \{0}
¢
¡
P
for all u ∈ Cc2 (Rd × Rd ) and with β = E0 ξ1 − 06s61 ∆ξτ 1{|∆ξs |>1} . In particular, the pairs
b of the process.
(u, Au), u ∈ Cc2 (Rd × Rd ), are in the full generator A
Proof. For u = u(x, p) ∈ Cc2 (Rd × Rd ) we can use Itô’s formula (for jump processes, now in the
usual form [30, p. 70, Theorem II.32]) and get with a similar calculation to the one made in the
proof of Theorem 3
u(Xt ,Pt ) − u(x0 , p0 ) =
+
1
2
Zt
tr
0
Ã
Zt
∂u
Ps ds −
∂x
0
∂2u
∂p2
Zt
0
µ
∂c
∂x
¶
Q
µ
∂c
∂x
∂u ∂V
ds −
∂p ∂x
¶T !
Zt
∂u ∂c
dξs
∂p ∂x
0
ds
¶
X µ
∂u
∂c
+
u(Xs , Ps ) − u(Xs , Ps− ) +
(Xs , Ps− )
∆ξs .
∂p
∂x
06s6t
Here we used the fact that the continuous martingale part of ξt is WtQ , and so [ξ, ξ]ct =
[W Q , W Q ]t = Qt. Note that we suppressed arguments in those places where no ambiguity
10
is possible. Since Ps = Ps− + ∆Ps = Ps− −
∂c
∂x
∆ξs we find, using the Lévy decomposition (3),
u(Xt , Pt ) − u(x0 , p0 )
=
Zt
∂u
Ps ds −
∂x
∂u ∂V
ds −
∂p ∂x
0
0
−
Zt
+
ZZ ³
∂u ∂c
∂p ∂x
0
+
Zt
ZZ ³
Z
Zt
∂u ∂c
β ds −
∂p ∂x
0
1
y Ñ (dy, ds) +
2
∂u ∂c
dWsQ
∂p ∂x
0
Zt
0
0<|y|<1
Zt
tr
Ã
∂2u
∂p2
µ
∂c
∂x
¶
Q
u(Xs , Ps− −
∂c
∂x
y) − u(Xs , Ps− ) +
∂u(Xs ,Ps− ) ∂c
∂p
∂x
u(Xs , Ps− −
∂c
∂x
y) − u(Xs , Ps− ) +
∂u(Xs ,Ps− ) ∂c
∂p
∂x
µ
∂c
∂x
¶T !
ds
´
y 1{|y|<1} Ñ (dy, ds)
´
y 1{|y|<1} ν(dy) ds
with the double integrals ranging over [0, t] × Rd \ {0}. The function u has compact support,
and we may take expectations on both sides of the above relation and differentiate in t. Since
the terms driven by Ñ (dy, ds) or dWsQ are martingales, we find
¯
¯
d
E (u(Xt , Pt ))¯¯
dt
t=0
∂u(x0 , p0 )
∂u(x0 , p0 ) ∂V (x0 ) ∂u(x0 , p0 ) ∂c(x0 )
=
p0 −
−
β
∂x
∂p
∂x
∂p
∂x
Ã
!
µ
¶ µ
¶
∂ 2 u(x0 , p0 ) ∂c(x0 )
∂c(x0 ) T
1
Q
+ tr
2
∂p2
∂x
∂x
Z ³
´
∂u(x0 ,p0 ) ∂c(x0 )
0)
y)
−
u(x
,
p
)
+
y
1
+
u(x0 , p0 − ∂c(x
0
0
{|y|<1} ν(dy),
∂x
∂p
∂x
Rd \{0}
which is what we claimed. Notice, that the convergence is pointwise, so that it is not clear
that Cc2 (Rd × Rd ) is in the domain of the generator. However, our calculation shows that
Au ∈ Cb (Rd × Rd ) and
Z t
E u(Xt , Pt ) − u(x0 , p0 ) =
E (Au)(Xs , Ps ) ds
0
b
which means that (u, Au) is in the full generator A.
If the driving Lévy process has no drift, no Brownian part and a rotationally symmetric Lévy
measure, the form of the infinitesimal generator becomes much simpler. In this case we have for
all u ∈ Cc2 (Rd × Rd )
∂u(x, p) ∂V (x)
∂u(x, p)
p−
∂x
∂p
∂x
Z ³
´
+ v.p.
u(x, p − ∂c(x)
y)
−
u(x,
p)
ν(dy),
∂x
Au(x, p) =
Rd
11
(17)
R
R
where v.p. Rd f (y) ν(dy) := limε→0 |y|>ε f (y) ν(dy) stands for the principal value integral. It is
not hard to see that
Z ³
´
u(x, p − ∂c(x)
v.p.
y)
−
u(x,
p)
ν(dy)
∂x
Rd
=
Z
Rd \{0}
³
u(x, p −
∂c(x)
∂x
y) − u(x, p) +
∂u(x,y) ∂c(x)
∂x
∂x
´
y 1{|y|<1} ν(dy)
or also
1
=
2
Z
Rd \{0}
³
u(x, p −
∂c(x)
∂x
y) + u(x, p +
∂c(x)
∂x
´
y) − 2u(x, p) ν(dy)
holds. The latter two representations do exist in the sense of ordinary integrals (just use a
simple Taylor expansion for u up to order two) and are frequently used in the literature. For
our purposes, formula (17) is better suited. Notice that all three representations extend A onto
C 2.
Theorem 6. Let d > 3, V ∈ C 2 (Rd ), c ∈ C 2 (Rd , Rd ) and {ξt }t>0 be a symmetric α-stable Lévy
process, 0 < α < 2. Then the process {(Xt , Pt )}t>0 solving (1) is transient.
Proof. We want to apply Lemma 4. Take the function
¡
¢−γ
uγ (x, p) = (H(x, p) − V0 )−γ = 12 p2 + V (x) − V0
with V0 = inf V − 1 and with a parameter γ > 0 which we will choose later. It is not hard to
see that for this u = uγ (x, p) and
n
o
D := (x, p) ∈ R2d : |x| + |p| 6 1 ,
a := 12 min uγ (x, p)
(x,p)∈D
conditions (ii), (iii) of Lemma 4 are satisfied.
Moreover, we have
∂uγ ∂V
∂uγ
p−
= 0.
∂x
∂p ∂x
Since {ξt }t>0 is a symmetric α-stable process, its Lévy measure is of the form ν(dy) =
cα |y|−d−α dy with cα given by (16), and (17) shows that
Z
¡ ¡
¢
¢ dy
∂c
.
Ãuγ (x, p) = cα v.p. uγ x, p + ∂x
y − uγ (x, p)
|y|d+α
Rd
We will see in Corollary 9 below (with B = ∂c/∂x and b = 2(V (x) − V0 )) that we can choose
γ > 0 in such a way that Ãuγ (x, p) 6 0. This, however, means that also condition (iv) of Lemma
4 is met.
Let χk ∈ Cc∞ (Rd ) be a cut-off function with 1Bk (0) 6 χk 6 1B2k (0) and set uk (x, p) :=
uγ (x, p)χk (x)χk (p). Clearly, uk ∈ Cc2 (Rd × Rd ) and we know from Lemma 5 that the pair
12
b The following considerations are close to those in [31]. Write
(uk , Auk ) is in the full generator A.
kgkA = kg1A k∞ . Using a Taylor expansion we find for some 0 < θ < 1 and all f ∈ C 2 (Rd × Rd )
¡
¢
∂c
f x, p + ∂x
y − f (x, p)
µ
¶ µ
¶
d
∂c
y) ∂c
1 X ∂ 2 f (x, p + θ ∂x
∂c
∂f (x, p) ∂c(x)
y+
y
y
=
∂p
∂x
2
∂pi ∂pj
∂x i ∂x j
i,j=1
and, therefore, for all compact sets K ⊂ Rd and (x, p) ∈ K × K,
¯
¯ Z
¯
¯
¢
¢
¡ ¡
¯v.p. f x, p + ∂c y − f (x, p) ν(dy)¯
∂x
¯
¯
Rd
¯
Z
¯
¯
6 ¯v.p.
|y|<1
¡ ¡
f x, p +
° °2
°
d4 °
° ∂c °
6
2 ° ∂x °K
Z
0<|y|<1
∂c
∂x y
¢
¯
Z
¯
¯
− f (x, p) ν(dy)¯ + 2
¢
ν(dy) kf kK×Rd
|y|>1
° 2 °
Z
°∂ f °
2
°
°
+2
|y| ν(dy) ° 2 °
∂p K×K̃
ν(dy) kf kK×Rd ,
|y|>1
where K̃ = K + {p ∈ Rd : |p| 6 k∂c/∂xkK }. Since the estimate of the local part in (17) is
obvious, we find
° °
° °
° 2 °
µ
¶
° ∂f °
° ∂f °
°∂ f °
°
°
°
kÃf kK×K 6 C kf kK×Rd + °
+°
+°
,
° ∂x °
° ∂p °
° ∂p2 °
K×K
K×K
K×K̃
for any f ∈ C 2 (Rd × Rd ) with kf kK×Rd < ∞ and with an absolute constant C = C(K, c, V )
depending only on K, k∂c/∂xkK and k∂V /∂xkK . Since p 7→ uγ (x, p) vanishes at infinity,
condition (i) of Lemma 4 is satisfied for the sequence (uk , Auk ) → (uγ , Ãuγ ).
The theorem follows now directly from Lemma 4.
Appendix
We will now give the somewhat technical proof that for some γ > 0 the function u γ (x, p) =
¢−γ
¡1 2
which we used in the proof of Theorem 6 satisfies condition (iv) of Lemma
2 p + V (x) − V0
4. We begin with a few elementary lemmas.
Recall that Euler’s Beta function B(x, y) is given by
B(x, y) =
Z1
tx−1 (1 − t)y−1 dt,
x, y > 0,
(18)
0
and satisfies the relations
B(x, y) = B(y, x)
and
B(x, y) =
13
x+y
B(x, y + 1),
y
(19)
cf. Gradshteyn and Ryzhik [9, §8.38]. A change of variable in (18) according to t = s 2 yields
B(x, y) =
Z1
1
(s2 )x− 2 (1 − s2 )y−1 ds,
x, y > 0.
−1
Lemma 7. For any v ∈ R \ {0}, a > 1, d > 3 we have
J(v) =
Z1
(1 − s2 )
d−3
2
ln(v 2 + 2vs + a) ds > ln(a) I d−3 .
(20)
2
−1
Proof. We observe that J(v) = J(−v) and
µ 2
¶
µ 2
¶
v
v
v
v
2
ln(v + 2vs + a) − ln(a) = ln
+ 2 s + 1 > ln
+2 s+1 .
a
a
a2
a
Therefore, we may assume that a = 1 and v > 0. Since J(0) = ln(a) = 0, it is enough to show
that J(v) is increasing. This is clear for v > 1 since v 7→ v 2 + 2vs + 1 increases for all parameter
values |s| 6 1. For 0 < v < 1 we calculate the derivative
J 0 (v) = 2
Z1
d−3
v+s
(1 − s2 ) 2 ds.
v 2 + 2vs + 1
−1
In the case d = 3 a few lines of simple calculations give
µ
¶ µ
¶
1
1+v
2
0
J (v) = 1 − 2 ln
+
v
1−v
v
which is clearly positive. If d > 3, we use the symmetry of the measure (1 − s2 )
0
J (v) =
Z1 µ
v+s
v−s
+ 2
2
v + 2vs + 1 v − 2vs + 1
−1
= 2v
Z1
¶
(1 − s2 )
d−3
2
d−3
2
ds
d−3
v 2 + 1 − 2s2
(1 − s2 ) 2 ds
2
2
2
2
(v + 1) − 4v s
−1
2v
= 2
(v + 1)2
Z1
−1
¶2j
∞ µ
X
d−3
2v
s2j (1 − s2 ) 2 ds,
(v + 1 − 2s )
2
v +1
2
2
j=0
since 2v(v 2 + 1)−1 6 1. The integrand can be written as
¶2j
∞ µ
X
2v
2
2
s2j
(v + 1 − 2s )
v2 + 1
j=0
¶2j
¶2j
∞ µ
∞ µ
X
X
2v
2v
2j
2
s −2
s2j+2
= (v + 1)
v2 + 1
v2 + 1
j=0
j=0
14
ds and find
= (v 2 + 1) +
∞
X
j=1
= (v 2 + 1) +
(
(v 2 + 1)
∞
2(v 2 − 1) X
v2 + 1
j=1
µ
µ
2v
2
v +1
2v
v2 + 1
¶2j
¶2j−2
− 1) 2 2(v 2 − 1)
s + 2
> (v + 1) + 2
v +1
v +1
2(v 2
2
−2
µ
µ
2v
2
v +1
¶2j−2 )
s2j
s2j
2v
2
v +1
¶2
s4
1 − s2
since v 2 − 1 6 0. This gives
 1
Z
Z1
d−3
2v
2(v 2 − 1)
0
2 d−3

J (v) > 2
(1 − s ) 2 ds + 2
s2 (1 − s2 ) 2 ds
2
v +1
(v + 1)
−1
−1

1
µ
¶2 Z
2
d−5
2v
2(v − 1)
+ 2
s4 (1 − s2 ) 2 ds
(v + 1)2 v 2 + 1
−1
¶
µ
2 − 1)
2v
8v 2
2(v
v2 − 1
3 d−1
5 d−3
1 d−1
= 2
B( 2 , 2 ) + 2
B( 2 , 2 ) .
B( 2 , 2 ) + 2
v +1
(v + 1)2
v + 1 (v 2 + 1)3
Using (19) we find for all dimensions d > 4
3 d−1
B( 12 , d−1
2 ) = dB( 2 , 2 )
B( 25 , d−3
2 )=
and
3
B( 32 , d−1
2 ),
d−3
and so
µ
2v
3 d−1
J (v) > 2
B( 2 , 2 ) d +
v +1
µ
2v
3 d−1
B( 2 , 2 ) 4 +
> 2
v +1
0
3 8v 2 (v 2 − 1)
2(v 2 − 1)
+
(v 2 + 1)2
d − 3 (v 2 + 1)4
¶
2(v 2 − 1) 24v 2 (v 2 − 1)
+
.
(v 2 + 1)2
(v 2 + 1)4
¶
It is now straightforward to check that
4+
2(v 2 − 1) 24v 2 (v 2 − 1)
+
>0
(v 2 + 1)2
(v 2 + 1)4
for all v ∈ R.
Lemma 8. Let d > 3, 0 < α < 2. There exists some γ = γ(α, d) > 0 such that
¶
Z µ
1
1
dy
v.p.
−
<0
2
γ
2
γ
(|p + λy| + 1)
(|p| + 1)
|y|d+α
Rd
holds for all p ∈ Rd , λ ∈ R.
15
(21)
Proof. With the reasoning following Lemma 5 it is clear that the integral (21) exists. Without
loss of generality we may assume that λ = 1. Denote the left-hand side of (21) by I(γ). Changing
to polar coordinates we get
ZZ
I(γ) =
Z(r) r
−1−α
drdθ = |S
d−2
|
Z∞
Z(r) r −1−α dr,
0+
S d−2 ×(0+,∞)
(in the sense of an improper integral at the lower limit 0+) where
Z(r) =
Z1 µ
1
1
−
2
2
γ
2
(r + |p| + 2r|p|s + 1)
(|p| + 1)γ
−1
¶
(1 − s2 )
¶
(1 − s2 )
d−3
2
ds.
d−3
2
ds.
Write Z(r) = |p|−2γ Z̃(r) and observe that with v = r/|p|
Z̃(r) =
Z1 µ
1
1
−
2
−2
γ
(v + 1 + 2vs + |p| )
(1 + |p|−2 )γ
−1
An application of Lemma 7 with a = 1 + |p|−2 implies
¯
Z1
¡
¢
d−3
∂ Z̃(r) ¯¯
=−
ln(v 2 + 2vs + a) − ln(a) (1 − s2 ) 2 ds
¯
∂γ ¯
γ=0
−1
³
´
= − J(v) − ln(a) I d−3 < 0,
2
and therefore
0
I (0) = −|p|
−α−2γ
Z∞ ³
0+
´
J(v) − ln(a)I d−3 v −1−α dv < 0.
2
Since I(0) = 0, the claim follows.
Assertion (iv) of Lemma 4 follows finally from
Corollary 9. Let d > 3 and 0 < α < 2. Then there exists some γ = γ(α, d) > 0 such that for
all B ∈ Rd×d , b > 0, p ∈ Rd
¶
Z µ
dy
1
1
6 0.
(22)
−
v.p.
(|p + By|2 + b)γ
(|p|2 + b)γ |y|d+α
Rd
Proof. An argument similar to the one used in the proof of Lemma 8 shows that the integral
(22) is well-defined for every γ > 0. Since
¶
Z µ
1
1
dy
v.p.
−
(|p + By|2 + b)γ
(|p|2 + b)γ |y|d+α
Rd
¶
Z µ
dy
1
1
1
− −1/2 2
= γ v.p.
,
−1/2
−1/2
2
γ
γ
b
|y|d+α
(|b
p+b
By| + 1)
(|b
p| + 1)
Rd
16
we may assume that b = 1. Depending on the rank of the matrix B we distinguish between
three cases.
Case 1: rank B = 0. Nothing is to prove in this case.
Case 2: rank B = d. We have
Z µ
J (λ) = v.p.
¶
dy
1
1
−
2
γ
2
γ
(|p + By| + 1)
(|p + λy| + 1)
|y|d+α
d
R
¶
Z µ
1
1
dy
α
= λ v.p.
−
−1
2
γ
2
γ
(|p + λ By| + 1)
(|p + y| + 1)
|y|d+α
Rd
and, therefore,
lim λ−α J (λ) < 0
and, by Lemma 8,
λ→0
lim λ−α J (λ) > 0.
λ→∞
Since J (λ) is a continuous function, there exists some λ∗ = λ∗ (p, B) such that J (λ∗ ) = 0. Thus,
¶
Z µ
1
dy
1
−
v.p.
(|p + By|2 + 1)γ
(|p + λy|2 + 1)γ |y|d+α
Rd
¶
Z µ
1
1
dy
∗
= J (λ ) + v.p.
6 0,
−
∗
2
γ
2
γ
(|p + λ y| + 1)
(|p| + 1)
|y|d+α
Rd
where we used Lemma 8 again.
Case 3: rank B = k, 1 < k < d. In this case we can find an orthogonal matrix S ∈ R d×d such
that
µ 0 ¶
B 0
B=S
ST
0 0
where B̃ ∈ Rk×k has full rank. Since the measure |y|−d−α dy µis invariant
under orthogonal
¶
0
B 0
transformations we can assume that B is already of the form
; otherwise we would
0 0
make a change of variables in (22) with p0 = Sp in place of p. Write y = (y1 , y2 ) ∈ Rk × Rd−k ,
p = (p1 , p2 ) ∈ Rk × Rd−k and set b = 1 + |p2 |2 . Then
¶
Z µ
1
1
dy
−
v.p.
2
γ
2
γ
(|p + By| + 1)
(|p + λy| + 1)
|y|d+α
d
R
¶
ZZ µ
1
dy1 dy2
1
= v.p.
−
d+α
0
2
γ
2
γ
(|p1 + B y1 | + b)
(|p1 | + b)
(|y1 |2 + |y2 |2 ) 2
Rd
¶ Z
Z µ
1
1
dy2
= v.p.
−
d+α dy1
2
(|p1 + B 0 y1 |2 + b)γ
(|p1 |2 + b)γ
(|y1 | + |y2 |2 ) 2
k
d−k
R
R
¶
Z
Z µ
dy1
dη2
1
1
=
−
d+α v.p.
0
2
γ
2
γ
(|p1 + B y1 | + b)
(|p1 | + b)
|y1 |k+α
(1 + |η2 |2 ) 2
d−k
k
R
R
17
where we used the change of variables |y1 |η2 = y2 in the last step. Since B 0 has full rank, the
claim follows from case 2.
Acknowledgement. Financial support for A. Tyukov through the NTU Research Enhancement Grant RF 175 and for R. Schilling by the Nuffield Foundation under grant NAL/00056/G
is gratefully acknowledged. Part of the work was done under EPSRC grant GR/R200892/01
supporting R. Schilling and A. Tyukov. We thank A. Truman and N. Jacob from the University
of Wales (Swansea) for their support and valuable discussions.
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