2.3. STABILITY 65
is,
Wt=W0+ Z t
0
Z 1 0
Z ∞ 0
Z 2π 0
c Ws−, Ws∗(α), z, ϕ+ϕ0(Xs−−Xs∗(α),Ws−−Ws∗(α))
×M(ds, dα, dz, dϕ), where theft-distributedWt∗ is optimally coupled withXt∗ for eacht ≥ 0. Unfortunately, we cannot prove that such a process exists, because of the term ϕ+ϕ0(Xs− −Xs∗(α), Ws− − Ws∗(α)). Such a problem was already encountered by Tanaka [57], and we more or less solve it as he did, by introducing, for allK ≥1,
WtK =W0+ Z t
0
Z 1 0
Z ∞ 0
Z 2π 0
cK(Ws−K, Ws∗(α), z, ϕ+ϕs,α,K)M(ds, dα, dz, dϕ) withϕs,α,K =ϕ0(Xs−−Xs∗(α), Ws−K −Ws∗(α))as a coupling SDE. This equation of course has a unique strong solution(WtK)t≥0, but the computation becomes more complicated.
Finally, we observe that
W22(ft,f˜t)≤lim sup
K→∞ E[|WtK−Xt|2], becauseWtK goes in law toftfor eacht≥0.
Using the Itô formula, we find
E[|WtK −Xt|2] =E[|W0−X0|2] +E Z t
0
Z 1 0
∆Ks (α)dαds
, where
∆Ks (α) :=
Z ∞ 0
Z 2π 0
|WsK−Xs+cK,W(s)−cX(s)|2− |WsK−Xs|2 dϕdz with the shortened notation
cK,W(s) :=cK WsK, Ws∗(α), z, ϕ+ϕs,α,K
and cX(s) :=c Xs, Xs∗(α), z, ϕ . Then we deduce from Section 2 that
∆Ks (α)≤C(|WsK−Xs|2+|Ws∗(α)−Xs∗(α)|2)|WsK−Ws∗(α)|γ
+C|WsK−Ws∗(α)|2+2γ/νK1−2/ν. It is then not too hard to conclude, using technical computations, that
lim sup
K→∞ E[|WtK−Xt|2]≤ W22(f0,f˜0) exp
Cγ,p
Z t 0
1 +kfskLp
ds
,
2.3. STABILITY 67 which completes the proof.
We first state the following result, of which the proof lies at the end of the section.
Proposition 2.3.1. Assume(2.3)for someγ ∈(−1,0),ν ∈(0,1)withγ+ν >0. Consider any weak solution ( ˜ft)t≥0 ∈ L∞ [0,∞),P2(R3)
to (2.1). Then there exists, on some probability space, a random variableX0with lawf˜0, independent of a Poisson measureM(ds, dα, dz, dϕ) on[0,∞)×[0,1]×[0,∞)×[0,2π)with intensitydsdαdzdϕ, a measurable family(Xt∗)t≥0of α-random variables such thatLα(Xt∗) = ˜ftand a càdlàg adapted process(Xt)t≥0 solving
Xt=X0+ Z t
0
Z 1 0
Z ∞ 0
Z 2π 0
c Xs−, Xs∗(α), z, ϕ
M(ds, dα, dz, dϕ) (2.21) and such that for allt≥0,L(Xt) = ˜ft.
We are unfortunately not able to say anything about uniqueness (in law) for this SDE, except iff˜is astrongsolution, and this is precisely the reason why things are complicated. We really need to use the ideas of [22] to produce, for( ˜ft)t≥0 given, a solution(Xt)t≥0 of which the time marginals are( ˜ft)t≥0.
Proposition 2.3.2. Assume (2.3) for some γ ∈ (−1,0), ν ∈ (0,1) with γ +ν > 0, that f0 ∈ Pq(R3)for some q ≥ 2such that q > γ2/(γ+ν)and that f0 has a finite entropy. Fix p∈ (3/(3 +γ), p0(γ, ν, q)). Let(ft)t≥0 ∈L∞ [0,∞),P2(R3)
∩L1loc [0,∞), Lp(R3) be the corresponding unique weak solution to(2.1)given by Theorem2.1.3. Consider also the Poisson measureM, the process(Xt)t≥0 and the family(Xt∗)t≥0 built in Proposition2.3.1(associated to another weak solution( ˜ft)t≥0 ∈L∞ [0,∞),P2(R3)
. LetW0 ∼ f0 (independent ofM) be such that E[|W0 −X0|2] = W22(f0,f˜0)and, for each t ≥ 0, an α-random variableWt∗ such thatLα(Wt∗) =ftandEα[|Wt∗−Xt∗|2] =W22(ft,f˜t). Then forK ≥1, the equation
WtK =W0 + Z t
0
Z 1 0
Z ∞ 0
Z 2π 0
cK(Ws−K, Ws∗(α), z, ϕ+ϕs,α,K)M(ds, dα, dz, dϕ), (2.22) with ϕs,α,K = ϕ0(Xs− −Xs∗(α), Ws−K −Ws∗(α)), has a unique solution. Moreover, setting ftK =L(WtK)for eacht≥0, it holds that for allT >0,
K→∞lim sup
[0,T]
W22(ftK, ft) = 0. (2.23) Remark 2.3.3. As recalled in the previous section, the infimum in the definition of Wasserstein distance is actually a minimum. Since the strong solution ft ∈ P2(R3) has a density for all t ≥ 0, there is a unique Rt ∈ H(ft,f˜t) such that W22(ft,f˜t) = R
R3×R3|v −v˜|2Rt(dv, d˜v) (see Villani [61, Theorem 2.12]). We then know that(t, α)7→ (Wt∗(α), Xt∗(α))can be chosen measurable from Fontbona-Guérin-Méléard [23, Theorem 1.3].
Proof. For any K ≥ 1, the Poisson measure involved in (2.22) is actually finite (because cK = c1{z≤K}), so the existence and uniqueness for this equation is obvious. It only remains to prove (2.23), which has already been done in [25, Lemma 4.2], where the formulation of the equation is slightly different. But one easily checks that(WtK)t≥0 is a (time-inhomogeneous) Markov process with the same generator as the one defined by [25, Eq. (4.1)], because for all bounded measurable functionφ :R3 7→Rand allt≥0, a.s.,
Z 1 0
Z ∞ 0
Z 2π 0
h
φ(w+cK(w, Wt∗(α), z, ϕ+ϕ0(Xt−−Xt∗(α), w−Wt∗(α)))−φ(w)i
dϕdzdα
= Z 1
0
Z ∞ 0
Z 2π 0
h
φ(w+cK(w, v, z, ϕ))−φ(w)i
dϕdzft(dv) by the2π-periodicity ofcK (inϕ) and sinceLα(Wt∗) =ft.
Now, we use these coupled processes to conclude the
Proof of Theorem2.1.4. We consider a weak solution( ˜ft)t≥0to (2.1), with which we associate the objectsM, (Xt)t≥0, (Xt∗)t≥0 as in Proposition 2.3.1. We then considerf0 satisfying the assumptions of Theorem2.1.3and the corresponding unique weak solution(ft)t≥0belonging to L∞ [0,∞),P2(R3)
∩L1loc [0,∞), Lp(R3)
(withp∈(3/(3+γ), p0(γ, ν, q))) and we consider (WtK)t≥0, (Wt∗)t≥0 built in Proposition 2.3.2 for any K ≥ 1. We know that W22(f0,f˜0) = E[|W0−X0|2]and thatW22(ft,f˜t) =Eα[|Wt∗−Xt∗|2]for allt≥0. Using thatWtK ∼ftK and Xt∼f˜tfor eacht≥0, we deduce from (2.23) that for allt≥0,
W22(ft,f˜t)≤lim sup
K→∞ E[|WtK−Xt|2] =: Jt. (2.24) Next, we focus on the time interval[0, T]for any fixedT > 0, and split the proof into several steps.
Step 1.By the Itô formula, we know that
E[|WtK −Xt|2] =E[|W0−X0|2] +E Z t
0
Z 1 0
∆Ks (α)dαds
, where
∆Ks (α) :=
Z ∞ 0
Z 2π 0
|WsK−Xs+cK,W(s)−cX(s)|2− |WsK−Xs|2 dϕdz with the shortened notation
cK,W(s) :=cK WsK, Ws∗(α), z, ϕ+ϕs,α,K
, cX(s) :=c Xs, Xs∗(α), z, ϕ .
2.3. STABILITY 69 We then show that
∆Ks (α)≤C(|WsK−Xs|2 +|Ws∗(α)−Xs∗(α)|2)|WsK−Ws∗(α)|γ
+C|WsK−Ws∗(α)|2+2γ/νK1−2/ν, (2.25) and
∆Ks (α)≤C|WsK−Ws∗(α)|γ+2+C|Xs−Xs∗(α)|γ+2
+C|WsK −Xs| |WsK−Ws∗(α)|γ+1+|Xs−Xs∗(α)|γ+1
. (2.26) First, Lemma2.2.3(inequality (2.18)) precisely tells us that (2.25) holds true. Next, we observe that
∆Ks (α)≤2 Z ∞
0
Z 2π 0
(|cK,W(s)|2+|cX(s)|2)dϕdz + 2|WsK−Xs|
Z ∞ 0
Z 2π 0
(cK,W(s)−cX(s))dϕdz . Hence, using (2.19) and (2.20), the proof of (2.26) is concluded.
Step 2. Set κ(γ) = min((γ + 1)/|γ|,|γ|/2) > 0. We verify that there exists a constant C(T, f0,f˜0, f) > 0(depending on T, m2(f0), m2( ˜f0), Rt
0kfskLpds), such that for all` ≥ 1 (and allK ≥1),
Iti,` ≤C(T, f0,f˜0, f)`−κ(γ), i= 1,2,3,4, where
It1,`:=E hZ t
0
Z 1 0
|WsK −Ws∗(α)|γ+21{|WsK−Ws∗(α)|γ≥`}dαdsi , It2,`:=E
hZ t 0
Z 1 0
|Xs−Xs∗(α)|γ+21{|WK
s −Ws∗(α)|γ≥`}dαdsi , It3,`:=E
hZ t 0
Z 1 0
|WsK −Xs||WsK−Ws∗(α)|γ+11{|WK
s −Ws∗(α)|γ≥`}dαdsi , It4,`:=E
hZ t 0
Z 1 0
|WsK −Xs||Xs−Xs∗(α)|γ+11{|WsK−Ws∗(α)|γ≥`}dαds i
. Sinceγ ∈(−1,0)andκ(γ)≤(γ + 2)/|γ|, we have
It1,`≤`−(γ+2)/|γ|T ≤`−κ(γ)T.
Similarly,
It3,` ≤`−(γ+1)/|γ|
Z t 0
E h
|WsK−Xs|i ds.
Using (2.9) for(ft)t≥0 and( ˜ft)t≥0, (2.23), and thatm2(fsK) ≤ 2m2(fs) + 2W22(fs, fsK), we know thatE
h|WsK −Xs|i
≤C(1 +m2(fsK) +m2( ˜fs))≤C(T, f0,f˜0). Hence, It3,` ≤C(T, f0,f˜0)`−κ(γ).
Sinceγ+ 2∈(1,2), it follows from the Hölder inequality that It2,`≤E
"
Z t 0
Z 1 0
|Xs−Xs∗(α)|2dαdsγ+22 Z t 0
Z 1 0
1{|WK
s −Ws∗(α)|γ≥`}dαds|γ|2
#
≤CE
"
Z t 0
(|Xs|2 +m2( ˜fs))dsγ+22 Z t 0
Z 1 0
|WsK−Ws∗(α)|γ
` dαds|γ|2
#
SinceLα(Ws∗) = fs, we haveR1
0 |WsK−Ws∗(α)|γdα =R
R3|WsK−v|γfs(dv)≤1 +Cγ,pkfskLp by (2.13), so that
It2,`≤`γ/2 1 +
Z t 0
E[|Xs|2] +m2( ˜fs) ds
Z t 0
1 +Cγ,pkfskLp
ds |γ|2
≤`γ/2
1 + 2m2( ˜f0)T 1 +
Z t 0
1 +Cγ,pkfskLp ds
≤C(T,f˜0, f)`−κ(γ). ForIt4,`, we use the triple Hölder inequality to write
It4,` ≤E hZ t
0
|WsK−Xs|2ds i12
×E hZ t
0
Z 1 0
|Xs−Xs∗(α)|2dαds i1+γ2
×E hZ t
0
Z 1 0
1{|WK
s −Ws∗(α)|γ≥`}dαdsi|γ|2 . ThusIt4,` ≤ C(T, f0,f˜0, f)`−κ(γ): use that E[|Xs|2] = Eα[|Xs∗|2] = m2( ˜f0), that m2(fsK) ≤ 2m2(fs) + 2W22(fs, fsK)as before and treat the last term of the product the same as we study It2,`.
Step 3. According to Step 1, we now bound∆Ks (α)by (2.25) when|WsK −Ws∗(α)|γ ≤ ` and by (2.26) when|WsK −Ws∗(α)|γ ≥`:
E[|WtK−Xt|2]
≤E[|W0 −X0|2] +C
4
X
i=1
Iti,`+CK1−2/νE hZ t
0
Z 1 0
|WsK−Ws∗(α)|2+2γ/νdαds i
+CE hZ t
0
Z 1 0
(|WsK−Xs|2+|Ws∗(α)−Xs∗(α)|2) min |WsK−Ws∗(α)|γ, ` dαds
i .
2.3. STABILITY 71 It then follows from Step 2 that for all` ≥1, allK ≥1,
E[|WtK−Xt|2]≤ W22(f0,f˜0) +C(T, f0,f˜0, f)`−κ(γ) (2.27) +CK1−2/νE
hZ t 0
Z 1 0
|WsK−Ws∗(α)|2+2γ/νdαdsi +CE
hZ t 0
Z 1 0
|WsK−Xs|2|WsK−Ws∗(α)|γdαds i
+CE hZ t
0
Z 1 0
|Ws∗(α)−Xs∗(α)|2min |WsK−Ws∗(α)|γ, ` dαdsi
. Sinceγ+ν > 0, it holds that2 + 2γ/ν >0. As a consequence, like in Step 2,
E hZ t
0
Z 1 0
|WsK−Ws∗(α)|2+2γ/νdαdsi
≤CT[1 +E[|WsK|2] +m2(f0)]≤C(T, f0,f˜0), which gives
K→∞lim K1−2/νE hZ t
0
Z 1 0
|WsK −Ws∗(α)|2+2γ/νdαdsi
= 0.
Moreover, we recall that a.s. R1
0 |WsK −Ws∗(α)|γdα≤1 +Cγ,pkfskLpas in Step 2, whence E
hZ t 0
Z 1 0
|WsK−Xs|2|WsK −Ws∗(α)|γdαds i
≤ Z t
0
E[|WsK−Xs|2](1 +Cγ,pkfskLp)ds.
LettingK → ∞, by dominated convergence, we find (recall (2.24)) lim sup
K E
hZ t 0
Z 1 0
|WsK−Xs|2|WsK −Ws∗(α)|γdαdsi
≤ Z t
0
Js(1 +Cγ,pkfskLp)ds.
Next, it is obvious that for each ` ≥ 1 fixed, for all s ∈ [0, T], all α ∈ [0,1], the func- tion v 7→ min(|v −Ws∗(α)|γ, `) is bounded and continuous. By (2.23), we conclude that limK→∞E
min |WsK−Ws∗(α)|γ, `
=E
min |Ws−Ws∗(α)|γ, `
and, by dominated con- vergence, that, still for`≥1fixed,
K→∞lim E hZ t
0
Z 1 0
|Ws∗(α)−Xs∗(α)|2min |WsK−Ws∗(α)|γ, ` dαdsi
= Z t
0
Z 1 0
|Ws∗(α)−Xs∗(α)|2E
min |Ws−Ws∗(α)|γ, ` dαds.
But since Ws ∼ fs, we have, for each α fixed, E[min (|Ws−Ws∗(α)|γ, `)] ≤ R
R3|Ws∗(α)− v|γfs(dv) ≤ 1 +Cγ,pkfskLp by (2.13). Furthermore, we have R1
0 |Ws∗(α)− Xs∗(α)|2dα =
Eα[|Ws∗−Xs∗|2] =W22(fs,f˜s)≤Js. All in all, we have checked that
K→∞lim E hZ t
0
Z 1 0
|Ws∗(α)−Xs∗(α)|2min |WsK−Ws∗(α)|γ, ` dαdsi
≤C Z t
0
Js(1 +kfskLp)ds.
Gathering all the previous estimates to letK → ∞in (2.27): for each`≥1fixed, Jt≤ W22(f0,f˜0) +C(T, f0,f˜0, f)`−κ(γ)+C
Z t 0
Js(1 +kfskLp)ds.
Letting now`→ ∞and using the Grönwall lemma, we find Jt≤ W22(f0,f˜0) exp
Cγ,p
Z t 0
1 +kfskLp ds
. SinceW22(ft,f˜t)≤Jt, this completes the proof.
It remains to prove Proposition2.3.1. We start with a technical result.
Lemma 2.3.4. Assume (2.3) for some γ ∈ (−1,0), some ν ∈ (0,1) with γ +ν > 0 and recall that the deviation functioncwas defined by(2.14). Considerf ∈ P2(R3)andφ(x) = (2π)−3/2e−|x|2/(2). Setf(w) = (f ∗φ)(w).
(i)There exists a constantC >0such that for allx∈R3, all∈(0,1), Z
R3
Z
R3
Z ∞ 0
Z 2π 0
|c(v, v∗, z, ϕ)|φ(v−x)
f(x) dϕdzf(dv)f(dv∗)≤C 1 +p
m2(f) +|x|
, (ii)For all ∈ (0,1), allR > 0, there is a constantCR, > 0(depending only onm2(f)) such that for allx, y ∈B(0, R),
Z
R3
Z
R3
Z ∞ 0
Z 2π 0
|c(v, v∗, z, ϕ)|
φ(v−x)
f(x) − φ(v−y) f(y)
dϕdzf(dv)f(dv∗)≤CR,|x−y|.
Proof. We start with (i) and setI(x) =R
R3
R
R3
R∞ 0
R2π
0 |c(v, v∗, z, ϕ)|φf(v−x)(x) dϕdzf(dv)f(dv∗).
Using (2.8) and (2.5), we see that|c(v, v∗, z, ϕ)| ≤ G(z/|v −v∗|γ)|v −v∗| ≤ C(1 +z/|v − v∗|γ)−1/ν|v−v∗|. Hence
I(x)≤C Z
R3
Z
R3
Z ∞ 0
(1 +z/|v−v∗|γ)−1/ν|v−v∗|φ(v−x)
f(x) dzf(dv)f(dv∗)
=C Z
R3
Z
R3
|v−v∗|1+γφ(v −x)
f(x) f(dv)f(dv∗).
2.3. STABILITY 73 Using now that|v−v∗|1+γ ≤1 +|v|+|v∗|, we find
I(x)≤C Z
R3
Z
R3
(1 +|v|+|v∗|)φ(v −x)
f(x) f(dv)f(dv∗)
≤C 1 +p
m2(f) + R
R3|v|φ(v−x)f(dv) f(x)
. To conclude the proof of (i), it remains to studyJ(x) = (f(x))−1R
R3|v|φ(v−x)f(dv). We introduceL := p
2m2(f), for which f(B(0, L))≥ 1/2(becausef(B(0, L)c)≤ m2(f)/L2).
Using that{v ∈R3 :|v| ≤2|x|+L} ∪ {v ∈R3 :|v−x| ≥ |x|+L}=R3, we write J(x) =
R
R3|v|φ(v−x)f(dv) R
R3φ(v−x)f(dv) ≤2|x|+L+ R
|v−x|≥|x|+L|v|φ(v−x)f(dv) R
|v−x|≤|x|+Lφ(v −x)f(dv) . Sinceφis radial and decreasing,
Z
|v−x|≥|x|+L
|v|φ(v−x)f(dv)≤φ(|x|+L)p m2(f) and
Z
|v−x|≤|x|+L
φ(v−x)f(dv)≥φ(|x|+L)f(B(x,|x|+L))≥φ(|x|+L)/2 owing to the fact thatB(0, L)⊂B(x,|x|+L). Hence,
J(x)≤2|x|+L+ 2p
m2(f)≤2|x|+ 4p m2(f) and this completes the proof of (i).
For point (ii), we set
∆(x, y) = Z
R3
Z
R3
Z ∞ 0
Z 2π 0
|c(v, v∗, z, ϕ)||F(x, v)−F(y, v)|dϕdzf(dv)f(dv∗), whereF(v, x) := (f(x))−1φ(v−x). Exactly as in point (i), we start with
∆(x, y)≤C Z
R3
Z
R3
|v−v∗|1+γ|F(v, x)−F(v, y)|f(dv)f(dv∗)
≤C Z
R3
(1 +p
m2(f) +|v|)|F(v, x)−F(v, y)|f(dv)
≤C|x−y|
Z
R3
(1 +p
m2(f) +|v|) sup
a∈B(0,R)
|OxF(v, a)|
f(dv)
for allx, y ∈B(0, R). But we have OxF(v, a) = 1
φ(v−a)R
R3(v−u)φ(u−a)f(du)
(f(a))2 . (2.28)
Indeed, recalling thatφ(x) = (2π)−3/2e−|x|2/(2), we observe that Oxφ(v−x) = 1
(v−x)φ(v −x)andOxf(x) = 1
Z
R3
φ(u−x)(u−x)f(du).
SinceF(v, a) := (f(a))−1φ(v−a), we have
OxF(v, a) = Oxφ(v−a)f(a)−φ(v−a)Oxf(a) (f(a))2
= φ(v−a)
(v−a)f(a)−R
R3φ(u−a)(u−a)f(du) (f(a))2
= φ(v−a)
R
R3φ(u−a)(v −a)f(du)−R
R3φ(u−a)(u−a)f(du)
(f(a))2 ,
whence (2.28). Using now thatJ(a) = (f(a))−1R
R3|u|φ(u−a)f(du)≤ 2|a|+ 4p m2(f) as proved in (i),
|OxF(v, a)| ≤ 1
φ(v−a) f(a)
R
R3(|v|+|u|)φ(u−a)f(du) f(a)
≤ 1
φ(v−a) f(a)
|v|+ 2|a|+ 4p
m2(f) . But we know thatφ(x)≤(2π)−3/2 and that
f(a)≥ Z
|v−a|≤|a|+L
φ(v−a)f(dv)≥φ(|a|+L)f(B(a,|a|+L))≥φ(|a|+L)/2 sinceB(0, L)⊂B(a,|a|+L). Hence,
sup
a∈B(0,R)
|OxF(v, a)| ≤ 2
e(R+L)2/(2)
|v|+ 2R+ 4p m2(f)
. Consequently, for allx, y ∈B(0, R),
∆(x, y)≤ 2C
e(R+L)2/(2)|x−y|
Z
R3
1 +p
m2(f) +|v| |v|+ 2R+ 4p
m2(f) f(dv)
≤CR,|x−y|,
whereCR, depends only onR, andm2(f)(recall thatL:=p
2m2(f)).
2.3. STABILITY 75 Finally, we end the section with the
Proof of Proposition2.3.1. We consider any given weak solution( ˜ft)t≥0 ∈L∞([0,∞),P2(R3)) to (2.1) and we write the proof in several steps.
Step 1. We introduce φ(x) = (2π)−3/2e−|x|2/(2) and f˜t(w) = ( ˜ft ∗φ)(w). For each t ≥0, we see thatf˜tis a positive smooth function. We claim that for anyψ ∈Lip(R3),
∂
∂t Z
R3
ψ(w) ˜ft(dw) = Z
R3
f˜t(dw) ˜At,ψ(w), where
A˜t,ψ(w) = Z
R3
Z
R3
Z ∞ 0
Z 2π 0
[ψ(w+c(v, v∗, z, ϕ))−ψ(w)]φ(v−w)
f˜t(w) dϕdzf˜t(dv∗) ˜ft(dv).
(2.29) Indeed,f˜t(w) =R
R3φ(v −w) ˜ft(dv)sinceφ(x)is even. According to (2.10) and (2.15), we have
∂
∂t
f˜t(w) = Z
R3
Z
R3
Z ∞ 0
Z 2π 0
[φ(v−w+c(v, v∗, z, ϕ))−φ(v−w)]dϕdzf˜t(dv∗) ˜ft(dv)
= Z
R3
Z K 0
Z 2π 0
Z
R3
φ(v−w+c(v, v∗, z, ϕ)) ˜ft(dv)−f˜t(w)
dϕdzf˜t(dv∗) +
Z
R3
Z
R3
Z ∞ K
Z 2π 0
[φ(v−w+c(v, v∗, z, ϕ))−φ(v−w)]dϕdzf˜t(dv∗) ˜ft(dv) for anyK ≥1. We thus have, for anyψ ∈Lip(R3),
∂
∂t Z
R3
ψ(w) ˜ft(dw)
= Z
R3
Z
R3
Z K 0
Z 2π 0
Z
R3
φ(v−w+c(v, v∗, z, ϕ))ψ(w) ˜ft(dv)dϕdzf˜t(dv∗)dw
− Z
R3
Z
R3
Z K 0
Z 2π 0
ψ(w) ˜ft(w)dϕdzf˜t(dv∗)dw +
Z
R3
Z
R3
Z
R3
Z ∞ K
Z 2π 0
[φ(v−w+c(v, v∗, z, ϕ))−φ(v−w)]ψ(w)dϕdzf˜t(dv∗) ˜ft(dv)dw.
Using the change of variablesw−c(v, v∗, z, ϕ)7→ w, we see that the first integral of the RHS equals
Z
R3
Z
R3
Z K 0
Z 2π 0
Z
R3
φ(v−w)ψ(w+c(v, v∗, z, ϕ)) ˜ft(dv)dϕdzf˜t(dv∗)dw.
Consequently,
∂
∂t Z
R3
ψ(w) ˜ft(dw)
= Z
R3
Z
R3
Z K 0
Z 2π 0
Z
R3
ψ(w+c(v, v∗, z, ϕ))φ(v−w) f˜t(w)
f˜t(dv)−ψ(w)
f˜t(w)dϕdzf˜t(dv∗)dw +
Z
R3
Z
R3
Z
R3
Z ∞ K
Z 2π 0
[φ(v−w+c(v, v∗, z, ϕ))−φ(v−w)]ψ(w)dϕdzf˜t(dv∗) ˜ft(dv)dw
= Z
R3
Z
R3
Z K 0
Z 2π 0
Z
R3
[ψ(w+c(v, v∗, z, ϕ))−ψ(w)]φ(v−w) f˜t(w)
f˜t(dv)dϕdzf˜t(dv∗) ˜ft(dw) +
Z
R3
Z
R3
Z
R3
Z ∞ K
Z 2π 0
[φ(v−w+c(v, v∗, z, ϕ))−φ(v−w)]ψ(w)dϕdzf˜t(dv∗) ˜ft(dv)dw.
LettingK increase to infinity, one easily ends the step.
Step 2. We setFt,(v, x) = ( ˜ft(x))−1φ(v −x). For a given X0 with lawf˜0, and a given independent Poisson measureN(ds, dv, dv∗, dz, dϕ, du)on[0,∞)×R3×R3×[0,∞)×[0,2π)×
[0,∞)with intensitydsf˜s(dv) ˜fs(dv∗)dzdϕdu, there exists a pathwise unique solution to Xt =X0+
Z t 0
Z
R3
Z
R3
Z ∞ 0
Z 2π 0
Z ∞ 0
c(v, v∗, z, ϕ)1{u≤Fs,(v,Xs− )}N(ds, dv, dv∗, dz, dϕ, du).
(2.30) This classically follows from Lemma 2.3.4, which precisely tells us that the coefficients of this equation satisfy someat most linear growth condition(point (i)) and somelocal Lipschitz condition(point (ii)).
Step 3.We now prove thatL(Xt) = ˜ftfor eacht≥0. We thus introducegt =L(Xt). By the Itô formula, we see that for allψ ∈Lip(R3),
∂
∂t Z
R3
ψ(w)gt(dw)
= Z
R3
gt(dw) Z
R3
Z
R3
Z ∞ 0
Z 2π 0
ψ(w+c(v, v∗, z, ϕ))−ψ(w)
Ft,(v, w)dϕdzf˜t(dv∗) ˜ft(dv)
= Z
R3
gt(dw) ˜At,ψ(w).
Thus( ˜ft)t≥0 and(gt)t≥0satisfy the same equation and we of course haveg0 = ˜f0 by construc- tion. The following uniqueness result allows us to conclude the step: for anyµ0 ∈ P2(R3), there exists at most one family(µt) ∈ L∞loc [0,∞),P2(R3)
such that for any ψ ∈ Lip(R3), anyt≥0,
Z
R3
ψ(w)µt(dw) = Z
R3
ψ(w)µ0(dw) + Z t
0
ds Z
R3
µs(dw) ˜As,ψ(w). (2.31)
2.3. STABILITY 77 This must be classical (as well as Step 2 is), but we find no precise reference and thus make use of martingale problems. A càdlàg adapted R3-valued process (Zt)t≥0 on some filtered probability space(Ω,F,Ft,P)is said to solve the martingale problemM P( ˜At,, µ0, Lip(R3)) ifP◦Z0 =µ0 and if for allψ ∈Lip(R3),(Mtψ,)t≥0is a(Ω,F,Ft,P)-martingale, where
Mtψ, =ψ(Zt)− Z t
0
A˜s,ψ(Zs)ds.
According to [10, Theorem 5.2] (see also [10, Remark 3.1, Theorem 5.1] and [38, Theorem B.1]), it suffices to check the following points to conclude the uniqueness for (2.31).
(i) there exists a countable family(ψk)k≥1 ⊂ Lip(R3)such that for allt ≥ 0, the closure (for the bounded pointwise convergence) of {(ψk,A˜t,ψk), k ≥ 1}contains {(ψ,A˜t,ψ), ψ ∈ Lip(R3)},
(ii) for eachw0 ∈R3, there exists a solution toM P( ˜At,, δw0, Lip(R3)), (iii) for eachw0 ∈R3, uniqueness (in law) holds forM P( ˜At,, δw0, Lip(R3)).
The fact that (2.30) has a pathwise unique solution proved in Step 2 (there we can of course replaceX0 by any deterministic pointw0 ∈ R3) immediately implies (ii) and (iii). Point (i) is very easy (recall that >0is fixed here).
Step 4. In this step, we check that the family((Xt)t≥0)>0 is tight inD([0,∞),R3). To do this, we use the Aldous criterion [1], see also [40, p 321], i.e. it suffices to prove that for all T > 0,
sup
∈(0,1)E sup
[0,T]
|Xt|
<∞, lim
δ→0 sup
∈(0,1)
sup
S,S0∈ST(δ)E
|XS0 −XS|
= 0, (2.32)
whereST(δ)is the set containing all pairs of stopping times(S, S0)satisfying0≤ S ≤ S0 ≤ S+δ≤T.
First, sinceXt ∼f˜t = ˜ft? φ, we haveE[|Xt|2] ≤2(m2( ˜ft) + 3)≤2m2( ˜f0) + 6. Thus for anyT >0, using Lemma2.3.4-(i),
E h
sup
[0,T]
|Xt|i
≤E
|X0| +E
hZ T 0
Z
R3
Z
R3
Z ∞ 0
Z 2π 0
|c(v, v∗, z, ϕ)|φ(v−Xs)
f˜s(Xs) dϕdzf˜s(dv) ˜fs(dv∗)dsi
≤E
|X0| +CE
Z T 0
(1 +|Xs|)ds
≤CT.
Furthermore, for anyT >0,δ >0and(S, S0)∈ ST(δ), using again Lemma2.3.4-(i), E
|XS0 −XS|
≤E
Z S+δ S
Z
R3
Z
R3
Z ∞ 0
Z 2π 0
|c(v, v∗, z, ϕ)|φ(v−Xs)
f˜s(Xs) dϕdzf˜s(dv) ˜fs(dv∗)ds
≤CE
Z S+δ S
(1 +|Xs|)ds
≤CE
"
δsup
[0,T]
(1 +|Xs|)
#
≤CTδ.
Hence (2.32) holds true and this completes the step.
Step 5. We thus can find some (Xt)t≥0 which is the limit in law (for the Skorokhod topology) of a sequence (Xtn)t≥0 with n & 0. Since L(Xtn) = ˜ftn by Step 3 and since f˜tn → f˜t by definition, we have L(Xt) = ˜ft for each t ≥ 0. It only remains to show that (Xt)t≥0 is a (weak) solution to (2.21). Using the theory of martingale problems, see Jacod [39, Theorem 13.55], it classically suffices to prove that for any ψ ∈ Cb1(R3), the process ψ(Xt)−ψ(X0)−Rt
0 Bsψ(Xs)dsis a martingale, where Btψ(x) =
Z 1 0
Z ∞ 0
Z 2π 0
ψ(x+c(x, Xt∗(α), z, ϕ))−ψ(x)
dϕdzdα.
But sinceLα(Xt∗) = ˜ft, this rewrites (recall (2.15)) Btψ(x) =
Z
R3
Z ∞ 0
Z 2π 0
ψ(x+c(x, v∗, z, ϕ)−ψ(x)
dϕdzf˜t(dv∗) = Z
R3
Aψ(x, v∗) ˜ft(dv∗).
We thus have to prove that for any0≤ s1 ≤... ≤sk ≤s ≤ t ≤ T, anyψ1, ..., ψk ∈Cb1(R3), and anyψ ∈Cb1(R3),
E[F(X)] = 0, whereF :D([0,∞),R3)7→Ris defined by
F(λ) = Yk
i=1
ψi(λsi)
ψ(λt)−ψ(λs)− Z t
s
Brψ(λr)dr . We of course start fromE[Fn(Xn)] = 0, where, recalling (2.29),
F(λ) =Yk
i=1
ψi(λsi)
ψ(λt)−ψ(λs)− Z t
s
A˜r,ψ(λr)dr . We then write
E[F(X)]
≤
E[F(X)]−E[F(Xn)]
+
E[F(Xn)]−E[Fn(Xn)]
.
2.3. STABILITY 79 On the one hand, we know from [24, Lemma 3.3] that(x, v∗) 7→ Aψ(x, v∗)is continuous on R3 ×R3 and bounded by C|x−v∗|γ+1. We thus easily deduce that F is continuous at each λ ∈ D([0,∞),R3) which does not jump at s1, ..., sk, s, t (this is a.s. the case of X ∈ D([0,∞),R3) because it has no deterministic time jump by the Aldous criterion). We also deduce that |F(λ)| ≤ C(1 +Rt
0
R
R3|λr−v∗|γ+1f˜r(dv∗)dr). Using that 0 < γ+ 1 < 1, that sup∈(0,1)E[sup[0,T]|Xt|] < ∞by Step 4 and recalling that Xn goes in law to X, we easily conclude that|E[F(X)]−E[F(Xn)]|tends to0asn → ∞.
On the other hand, since|F(λ)− F(λ)| ≤C|Rt
s(Brψ(λr)−A˜r,ψ(λr))dr|andXr ∼f˜r,
E[F(Xn)]−E[Fn(Xn)]
≤C Z t
s
E h
Z
R3
Z ∞ 0
Z 2π 0
Z
R3
ψ(Xrn+c(v, v∗, z, ϕ)) hφn(v−Xrn)
f˜rn(Xrn)
f˜r(dv)−δXrn(dv)i
dϕdzf˜r(dv∗) i
dr
=C Z t
s
Z
R3
Z ∞ 0
Z 2π 0
Z
R3
Z
R3
ψ(w+c(v, v∗, z, ϕ)) h
φn(v−w) ˜fr(dv)−f˜rn(w)δw(dv)i
dwdϕdzf˜r(dv∗) dr.
But we can write Z
R3
Z
R3
ψ(w+c(v, v∗, z, ϕ)) ˜frn(w)δw(dv)dw
= Z
R3
ψ(w+c(w, v∗, z, ϕ)) ˜frn(w)dw
= Z
R3
Z
R3
ψ(w+c(w, v∗, z, ϕ))φn(v−w) ˜fr(dv)dw, so that
E[F(Xn)]−E[Fn(Xn)]
≤C Z t
s
Z
R3
Z ∞ 0
Z 2π 0
Z
R3
Z
R3
h
ψ(w+c(v, v∗, z, ϕ))−ψ(w+c(w, v∗, z, ϕ))i φn(v−w) ˜fr(dv)dwdϕdzf˜r(dv∗)
dr
=C Z t
s
Z
R3
Z ∞ 0
Z 2π 0
Z
R3
Z
R3
h
ψ(w+c(v, v∗, z, ϕ+ϕ0(v−v∗, w−v∗)))
−ψ(w+c(w, v∗, z, ϕ))i
φn(v−w) ˜fr(dv)dwdϕdzf˜r(dv∗) dr.
The last equality uses the2π-periodicity ofc. We now put Rn(v, v∗, z, ϕ) :=
Z
R3
h
ψ(w+c(v, v∗, z, ϕ+ϕ0(v−v∗, w−v∗)))
−ψ(w+c(w, v∗, z, ϕ))i
φn(v−w)dw, and show the following two things:
(a) for allv, v∗ ∈R3, allz ∈[0,∞)andϕ∈[0,2π),limn→∞Rn(v, v∗, z, ϕ) = 0;
(b) there is a constant C > 0such that for all n ≥ 1, all v, v∗ ∈ R3, all z ∈ [0,∞) and ϕ∈[0,2π),
|Rn(v, v∗, z, ϕ)| ≤C 1 +|v−v∗|
(1 +z)−1/ν,
which belongs to L1([0, T] × R3 × R3 × [0,∞) × [0,2π), drf˜r(dv∗) ˜fr(dv)dzdϕ) because ( ˜ft)t≥0 ∈L∞([0, T],P2(R3))by assumption.
By dominated convergence, we will deduce thatlimn→∞
E[F(Xn)]−E[Fn(Xn)]
= 0 and this will conclude the proof.
We first study (a). Sinceψ ∈Cb1(R3), we immediately observe that
ψ(w+c(v, v∗, z, ϕ+ϕ0(v−v∗, w−v∗)))−ψ(w+c(w, v∗, z, ϕ))
(2.33)
≤Cψ
c(v, v∗, z, ϕ+ϕ0(v −v∗, w−v∗))−c(w, v∗, z, ϕ) . Recalling that
c(v, v∗, z, ϕ) =−1−cosG(z/|v−v∗|γ)
2 (v−v∗) + sinG(z/|v−v∗|γ))
2 Γ(v−v∗, ϕ), we have
c(v, v∗, z, ϕ+ϕ0(v−v∗, w−v∗))−c(w, v∗, z, ϕ)
≤|cosG(z/|v−v∗|γ)−cosG(z/|w−v∗|γ)|
2 |v−v∗|+|1−cosG(z/|w−v∗|γ)|
2 |v−w|
+ |sinG(z/|v −v∗|γ)−sinG(z/|w−v∗|γ)|
2 |Γ(v−v∗, ϕ+ϕ0)|
+ |sinG(z/|w−v∗|γ)|
2 |Γ(v−v∗, ϕ+ϕ0)−Γ(w−v∗, ϕ)|.
Using that|Γ(v−v∗, ϕ+ϕ0)|=|v−v∗|and Lemma2.2.2, we obtain
c(v, v∗, z, ϕ+ϕ0(v−v∗, w−v∗))−c(w, v∗, z, ϕ)
≤C|G(z/|v−v∗|γ)−G(z/|w−v∗|γ)||v−v∗|+C|v−w|.
2.3. STABILITY 81 We deduce from (2.4) that|G0(z)|= 1/β(G(z))≤Cby (2.3), whence
c(v, v∗, z, ϕ+ϕ0(v−v∗, w−v∗))−c(w, v∗, z, ϕ)
≤Cz
|v−v∗||γ|− |w−v∗||γ|
|v−v∗|+C|v−w|.
Using again the inequality|xα−yα| ≤ |x−y|(x∨y)α−1forα∈(0,1), andx, y ≥0, we have |v−v∗||γ|− |w−v∗||γ|
≤ |v−w||v−v∗||γ|−1. We thus get
c(v, v∗, z, ϕ+ϕ0(v−v∗, w−v∗))−c(w, v∗, z, ϕ)
≤C(z|v−v∗||γ|+ 1)|v−w|.
Consequently,
Rn(v, v∗, z, ϕ)≤Cψ(z|v−v∗||γ|+ 1) Z
R3
|v−w|φn(v−w)dw, which clearly tends to0asn → ∞. This ends the proof of (a).
For (b), start again from (2.33) to write
ψ(w+c(v, v∗, z, ϕ+ϕ0(v−v∗, w−v∗)))−ψ(w+c(w, v∗, z, ϕ))
≤
ψ(w+c(v, v∗, z, ϕ))−ψ(w) +
ψ(w)−ψ(w+c(w, v∗, z, ϕ))
≤Cψ(|c(v, v∗, z, ϕ)|+|c(w, v∗, z, ϕ)|).
Moreover, since|c(v, v∗, z, ϕ)| ≤G(z/|v−v∗|γ)|v−v∗| ≤C|v−v∗|(1 +|v−v∗||γ|z)−1/νby (2.8) and (2.5), we observe that
Rn(v, v∗, z, ϕ)≤C|v−v∗|(1 +|v −v∗||γ|z)−1/ν +C
Z
R3
|w−v∗|(1 +|w−v∗||γ|z)−1/νφn(v−w)dw.
Since1 +|v−v∗||γ|z ≥ 1∧ |v −v∗||γ|
(1 +z)forz ∈[0,∞),
|v−v∗|(1 +|v−v∗||γ|z)−1/ν ≤ |v−v∗|(1 +z)−1/ν 1∧ |v −v∗||γ|−1/ν . Using that|γ|/ν <1, we deduce that
|v−v∗|(1 +|v−v∗||γ|z)−1/ν ≤ 1 +|v−v∗|
(1 +z)−1/ν. As a conclusion,
Rn(v, v∗, z, ϕ)≤C
1 +|v−v∗|+ Z
R3
|w−v∗|φn(v−w)dw
(1 +z)−1/ν, which is easily bounded (recall thatn ∈(0,1)) byC(1 +|v|+|v∗|)(1 +z)−1/νas desired.