We consider the functionu1, which is the maximal solution on(1,∞)of the non linear differential equation
u00(r)+d−1
r u0(r)=4u(r)2 .
Lemma 5.1. There exist positive constantsa0,b0andb01, depending only ond, such that
r→∞lim rd−2u1(r)=a0 ifd ≥5, lim
r→∞r2log(r) u1(r)=a0=1/2 ifd =4;
furthermore for everyr >1,
u1(r)≥a0r2−d ifd ≥5, u1(r)≥a0r−2log(2r)−1 ifd =4; (24) and for everyr ≥4/3,
u1(r)≤b0r2−d ifd ≥5, u1(r)≤b0[2r2log(r)]−1 ifd =4; (25) u1(r)≤a0r2−d+b01r6−2d ifd ≥5, (26) u1(r)≤a0r−2log(r)−1+b01r−2log(r)−2log(log(r)) ifd =4 .(27) Ford ≥5, we will see the constant a0can be expressed as the radius of convergence of a series. We will prove this lemma by giving the asymptotic expansion ofu1at∞.
Lemma 5.2. Ifd ≥5, we have u1(r)=r2−d
X∞ n=0
anr−n(d−4), r >1 ,
wherea0 is as in the above lemma and the sequence(an) is given by the recurrence:
an= 4
nδ(nδ+1)(d−2)−2 Xn k=0
akan−k−1, for n≥1
andδ= d−4 d−2. Ford =4, we have
u1(r)= 1 r2
1
2 log(r)+log(log(r))
4 log(r)2 +O log(r)−2
at + ∞ .
We introduce the auxiliary function z(t)=4(d−2)−2(d−1)/(d−2)t u1
" t d−2
1/(d−2)#
, fort > d −2 . This function is a positive solution on(d−2,∞)of
y00(t)=t−δ−2y(t)2 , (28) Let η > 0 be fixed. Set s = t −d +2 −η, z(s)˜ = z(t) and φ(s) = (s+d−2+η)−δ−2. Thenz˜solvesy00(s)=φ(s)y(s)2,s ≥0. We deduce from [18] p.132 case I (takeσ = −δ −2, λ = 2) that the functionz˜ is decreasing for s ≥ 0. Since η > 0 is arbitrary, we get that z itself (i.e.
rd−2u1(r)) is decreasing.
Proof of Lemma 5.2 in the case d ≥ 5. We deduce from theorems 1.1 and 2.4 of [18] (see also p.132 case 3, wherea > 0 is implicit) that the limit q = lims→∞z(s)˜ = limt→∞z(t) exists and is positive. Hence by integrating (28) twice fromt to∞, we get fort > d−2,
z(t)−q = Z ∞
t (r−t)r−δ−2z(r)2dr . (29)
Now consider the sequence(qn, n≥0)defined byq0=1 and the recurrence
qn = 1
nδ(nδ+1) Xn−1 k=0
qkqn−k−1, for n≥1 .
Clearly we have for everyn≥ 0,qn ≤ 2 [4/δ]nγn+1,where the sequence (γn, n ≥ 1)is introduced in the appendix. Thus the radius of convergence R of the seriesP
qnsn is bounded from below byδ/4. The power series z0(t) = P
qnqn+1t−δn is convergent and even C∞ as a function oft for t > t1=[q/R]1/δ. This power series also solves (29) fort > t1. The same arguments as in the proof of the Gronwall lemma show that equation (29) possesses a unique solution bounded in a neighborhood of infinity. Thus the functionszandz0agree fort > t1∨(d−2).
Since limt↓d−2z(t) = +∞, we gett1 ≤ d−2. Let us now prove that t1 ≥ d −2. Since q and the coefficientsqn are positive, it is enough to prove that for any integer p, z(t) ≥ vp(t) for t ∈ (d −2,+∞), where vp(t)=Pp
n=0qnqn+1t−δn, and then letpgoes to infinity to gett1 ≥d−2.
We consider the functionf = z−vp defined on(d −2,+∞). We have f >0 at least overI =(d−2, d−2+η)∪(η−1,+∞), forηsmall. It is easy to check, using the definition ofqnthatf00(t) ≥ t−δ−2[z(t)+vp(t)]f (t).
Hencef is convex whenf is positive. If there existstsuch thatf (t)≤ 0, then sincef is positive onI, there exists a last zerot0off that isf (t0)=0
andf > 0 on J = (t0,+∞). Now f is convex and positive on J, and f (t0) = limt→+∞f (t) = 0. This is absurd. Thus we deduce that f is positive over(d−2,+∞). As we noticed this in turn implies thatt1=d−2.
The radius of convergence of the seriesP
qnsnisq(d−2)−δ and we have fort > d−2
z(t)= X∞ n=0
qnqn+1t−nδ .
Thus we get with obvious notation forr >1, u1(r)=4−1(d−2)d/(d−2)r2−d
X∞ n=0
qnqn+1(d −2)−n(d−4)/(d−2)r−n(d−4)
=r2−d X∞
n=0
anr−n(d−4) .
The recurrence formula for(an)is a consequence of the recurrence formula for(qn).
Proof of Lemma 5.1 (d ≥ 5). From the above expression we easily deduce (25) and (26). Since the real numbers(an, n≥0)are positive, (24) follows easily. Notice that 4(d −2)−2a0 is the radius of convergence of the series Pqnsn.
Proof of Lemma 5.2 in the cased =4.We writef (t)∼g(t)at 0+when the real functionf and gare positive or negative onI = (0,0+ε) for someε > 0 and limt∈I,t→0f (t)/g(t) = 1. We also writef (t) ∼ g(t)at
∞whenf (1/t)∼ g(1/t)at 0+. Sincez ≥ 0, we know from [18] p.133 case 4, thatz(t) ∼ log(t)−1at ∞. We deduce from (28) that zis convex positive and limt→+∞z(t) = 0. This impliesz0(t)is negative on(2,∞).
From (28), we also havez00(t)∼[tlog(t)]−2at∞. By integration, we get z0(t) ∼ t−1log(t)−2 at ∞. We now consider the function w(s) = z(es) which solvesw00−w0 = w2on (log 2,∞). Notice that the functionw is positive decreasing andw0is negative. We also havew(s)∼s−1,w0(s)∼
−s−2andw00(s)=o(s−2)at∞. Thus the function defined on(0,∞)by p(w(s))=w0(s), for s ∈(log 2,∞) ,
is well defined and even of classC1, andp0(w(s))=w00(s)/w0(s). Thus the functionpcan be extended as aC1function on [0,∞)by settingp(0)=0 andp0(0)=0. Furthermore it solves
p(w)p0(w)−p(w)=w2 on [0,∞) .
We also havep(w) ∼ −w2 at 0+. We consider the sequence(ρn, n ≥ 2) defined byρ2=1 and the recurrence
ρn= Xn−1 k=2
kρkρn−k+1, for n≥3 . The radius of convergence of the seriesP
(−1)n+1ρnwnis 0, nevertheless we will prove this is the asymptotic expansion ofpat 0+. We setHn(w)= Pn
k=2(−1)k+1ρkwk for n ≥ 2. We now prove by induction that p(w) = Hn(w)+hn(w), wherehn(w)=o(wn)at 0+. This is true forn=2. Let us assume it is true at stagen. Letgn,α(w)=(1−α)(−1)nρn+1wn+1−hn(w).
We easily have
g0n,α(w)p(w)+gn,α(w)[Hn0(w)−1]
=
α(−1)nρn+1wn+1+o(wn+1),
(−1)n+1ρn+2wn+2+o(wn+2), ifα =0.
Let us assumenis even. Forα =0, the above right hand side is negative on (0, ε], for ε small enough. Sincep is negative and [Hn0(w)−1] < 0 on [0, ε], for ε small, we see that gn,0(w) < 0 impliesgn,00 (w) ≥ 0. As gn,0(0) =0, we get by contradiction thatgn,0 ≥ 0 on [0, ε]. This implies hn(w) ≤ ρn+1wn+1. Similar arguments for α > 0 implies that gn,α ≤ 0 on [0, εα] for εα > 0 small enough. Since this holds for any α > 0 and sincehn(w)≤ ρn+1wn+1forwsmall enough, we deduce thathn+1(w)= hn(w)−ρn+1wn+1=o(wn+1). Ifnis odd the proof is similar.
From the definition ofp, we then havew0(s)=Hn(w(s))+O(w(s)n+1) at∞. Forn =3 this givesw0(s)= −w(s)2+2w(s)3+O(w(s)4)at∞.
Sincew(s)∼s−1at+∞, we deduce by integration that 1
w(s)−2 logw(s)+O(1)=s at infinity .
Standard arguments yieldsw(s)=s−1+2s−2log(s)+O(s−2)at infinity.
Thus we have u1(r)= 1
r2 1
2 log(r)+log(log(r))
4 log(r)2 +O log(r)−2
at + ∞ . Notice the previous calculation can be continued to give an asymptotic expansion ofu1.
Proof of Lemma 5.1 (d = 4). The inequalities (25) and (27) follow easily from the above equality. We will now prove that for everyr >1,u1(r) ≥ [2r2log(2r)]−1. We consider the functionf (r)=u1(r)−[2r2log(2r)]−1.
The functionf is positive at least over(1,1+η)∩(η−1,∞)forηsmall. Let us assume thatf achieves its minimum atr0and thatf (r0) ≤0. Then we haver0 ∈ [1+η, η−1],f0(r0)=0 andf00(r0) ≥0. An easy computation gives
f00(r)=4f (r)
u1(r)+ 1 2r2log(2r)
−3
rf0(r)− 1
2r4(log(2r))3 . Evaluation atr =r0implies thatf00(r0) <0. This contradicts the assump-tion. Hencef is positive, that is we get (24) ford =4.
6. Appendix
6.1. Formula for moments of the Brownian snake
For the reader’s convenience, we recall some explicit formulas for moments of the Brownian snake. These formulas are well-known, at least in the con-text of superprocesses (see e.g. Dynkin [5]). We can compute the Laplace functional ofRt
0 ds(Ys, ϕ(s))forϕ ∈Bb+(R+×Rd). To this end start from the finite dimensional Laplace functional (2) withti =i/m,ϕi= m1 ϕ(i/m) for a nonnegative continuous functionϕwith compact support onR+×Rd. Thanks to the continuity of the processX, by a suitable passage to the limit, we get forν∈Mf
EXν
exp
− Z t
0 (Xt−s, ϕ(s)) ds
=exp [−(ν, v(t))] ,
where v is a nonnegative solution of (1) with right-hand side J (t, x) = Rt
0 ds Pt−s[ϕ(s)](x). This can be extended by monotone class arguments to anyϕ ∈Bb+(R+×Rd). The uniqueness of the solution is easily established using arguments similar to the classical Gronwall lemma. Then we get v(t, x)=Nx[1−exp[−Rt
0 ds (Yt−s, ϕ(s))]], thanks to theorem 1.3.
Now we introduce an auxiliary power series. Let us consider the analytic functionf (λ)=1−√
1−λfor|λ|<1. It is easy to check that for|λ|<1, we have
f (λ)= X∞
n=1
γnλn ,
where the sequence(γn, n≥1)is defined byγ1=1/2 and the recurrence γn = 1
2 Xn−1 k=1
γkγn−k for n≥2
(use the fact thatf solves 2f (λ)=f (λ)2+λ). Now letT >0 andJa non-negative measurable function onR+×Rd, such thatMT=sup[0,T]×RdJ(t, x)
<∞. We define the family of measurable functions(hn, n≥ 1)onR+× Rd, by the initial condition
h1(t)=J (t) , and the recurrence
hn(t)=2 Xn−1
k=1
Z t
0 ds Ps
hk(t−s)hn−k(t−s)
for n≥2 . (30) We clearly have for everyn≥1,
sup
[0,T]×Rd
|hn| ≤[4T]n−1[2MT]nγn . Thus the power seriesw(λ, t)=P
(−1)n+1λnhn(t)is normally convergent on [0, T] ×Rd for |λ| < [8T MT]−1. And it clearly solves the integral equation on [0, T]×Rd
w(t)+2 Z t
0 ds Ps
w(t −s)2
=λJ (t) . (31)
To get the uniqueness of the solution to the previous integral equation, use ar-guments similar to Gronwall’s lemma. Finally we can compute the moments for the processYunderNx. Indeed, letϕ ∈Bb+(R+×Rd). We have shown that forλ >0, the functionvλ(t, x)=Nx
h1−exp−λRt
0(Yt−s, ϕ(s)) dsi is the unique solution to (31) onR+×RdwithJ (t, x)=Rt
0 ds Ps[ϕ(t− s)](x). Thus forλ≥0 small enough, we havevλ(t)=w(λ, t). Then from the series expansion forw(λ, t), we get for every integern≥1
Nx
Z t
0 ds (Yt−s, ϕ(s)) n
=n!hn(t, x) , where the functions hn are defined by h1(t) = Rt
0 ds Ps[ϕ(t −s)], and the recurrence (30). In the same way it can be shown that for every ϕ ∈ Bb+(Rd), for everyt≥0,n≥1,
Nx
(Yt, ϕ)n
=n!hn(t, x) ,
where the functions are defined byh1(t)=Pt[ϕ], and the recurrence (30).
6.2. Proof of Proposition 1.4
We use the notations of the first section. Let Lthe last exit time ofA = w∈W∗x0;τB(x,ε)¯ (w) <∞ :
L=sup
s >0; ∃t ∈[0, ζs], Ws(t)∈ ¯B(x, ε) ,
with the convention sup∅ = 0. It is enough to prove the proposition with T(x,ε) replaced byL, sinceWLandWT(x,ε) are identically distributed under Nx0, thanks to the invariance ofNx0 under time reversal. We now compute forλ≥0,t ≥0,
Nx0
e−λL◦θtF (Wt, ζt)1L>t ,
whereF is a bounded nonnegative continuous function onW∗x0which van-ished on a neighborhood ofx0. We refer to [11] for the new notations and the properties of the Brownian snake used here. We often writeζ forζ(w). Using the Markov property at timet, we deduce from the characterization ofNx0 that
Nx0
e−λL◦θtF (Wt, ζt)1L>t
= Z
Mx0(dw)qt(ζ )F (w, ζ)E∗w
e−λL1L>0 , whereMx0(dw)=R∞
0 daPax0(dw)andPax0(dw)is the law ofβ[0,a]started at x0, viewed as a probability measure onWx0. Hereqt(a)=(2πt3)−1/2aexp [−a2/2t] is the entrance density under Itô measure of positive excursions.
We have E∗w
e−λL1L>0
=E∗w[1L>0]− Z ∞
0 ds λe−λsE∗w[1L>s] . We first consider
Nλ = Z
Mx0(dw)qt(ζ)F (w, ζ) Z ∞
0 ds λe−λsE∗w[1L>s] . LetQ∗s(w, dw0)denote the transition kernel of the Brownian snake killed when its lifetime reaches 0. Then we have
Nλ = Z ∞
0 ds λe−λs Z
Mx0(dw)qt(ζ)F (w, ζ ) Z
Q∗s(w, dw0)E∗w0[1L>0] . Recall that (Ws,E∗w) is symmetric with respect to the reference measure Mx0(dw), cf [11]. Thus the measure Mx0(dw)Q∗s(w, dw0) is symmetric, and we have
Nλ = Z ∞
0 ds λe−λs Z
Mx0(dw) Z
Q∗s(w, dw0)qt(ζ0)F (w0, ζ0)E∗w[1L>0] .
Let U(w00, dw) denotes the potential kernel of (Ws,E∗w). Using the def-inition of the capacitary measure of A (that is R
πA(dw00)U(w00, dw) = Mx0(dw)E∗w[1L>0]), we get
Nλ = Z ∞
0 ds λe−λs Z
πA(dw00) Z
U(w00, dw)
× Z
Q∗s(w, dw0)qt(ζ0)F (w0, ζ0)
= Z ∞
0 ds λe−λs Z
πA(dw00) Z ∞
s dr
Z
Q∗r(w00, dw0)qt(ζ0)F (w0, ζ0)
= Z ∞
0 dr
1−e−λr Z
πA(dw00) Z
Q∗r(w00, dw0)qt(ζ0)F (w0, ζ0) . By lettingλ→ ∞, we also have
Z
Mx0(dw)qt(ζ)F (w, ζ)E∗w[1L>0]
= Z ∞
0 dr Z
πA(dw00) Z
Q∗r(w00, dw0)qt(ζ0)F (w0, ζ0) . Thus we get
Nx0
e−λL◦θtF (Wt, ζt)1L>t
= Z ∞
0 dr e−λr Z
πA(dw) Z
Q∗r(w, dw0)qt(ζ0)F (w0, ζ0) . The end of the proof is now similar to the end of the proof of theorem 3.2.1 from Port and Stone [17] concerning the interpretation of the capacitary measure as a last exit distribution (use thatR∞
0 qt(a) dt =1).
Acknowledgements. I would like to thank my advisor J.-F. Le Gall for his help and advices.
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