Continuous time random walks and the Cauchy problem for the heat equation

0.2 MB·English

by  Hugo Aimar

CONTINUOUS TIME RANDOM WALKS AND THE CAUCHY
PROBLEM FOR THE HEAT EQUATION
6
1
0 HUGOAIMAR,GASTO´NBELTRITTI,ANDIVANAGO´MEZ
2

y
Abstract. Inthispaperwedealwithanomalous diffusionsinducedbyCon-
a
M tinuous TimeRandomWalks-CTRWinRn. AparticlemovesinRn insuch
a way that the probability density function u(·,t) of finding itin regionΩ of
6 Rn is given by RΩu(x,t)dx. The dynamics of the diffusion is provided by a
space time probability density J(x,t) compactly supported in {t≥0}. For t
2
largeenough,umustsatisfytheequationu(x,t)=[(J−δ)∗u](x,t)whereδis

theDiracdeltainspacetime. WegiveasensetoaCauchytypeproblemfora
]
P giveninitialdensitydistributionf. WeuseBanachfixedpointmethodtosolve
it,andweprovethatunderparabolicrescalingofJ theequationtendsweakly
A
totheheatequationandthatforparticularkernelsJ thesolutionstendtothe
. correspondingtemperatures whenthescalingparameterapproaches tozero.
h
t
a
m
[ 1. Introduction and statement of the results

2 Weshallbeconcernedwithaprobabilisticdescriptionofthemotionofaparticle
v in the space Rn. As usual we shall write Rn+1 to denote the set {(x,t) : x ∈
7 +
Rn and t ≥ 0}. Sometimes we shall also considerer the whole space time Rn+1 =
2
1 {(x,t) : x ∈ Rn and t ∈ R}. The x variable is thought as a space variable, while t
2 represents time.
0 Let u(x,t) denote, for t fixed, the probability density of the position of the
1. particle at time t. Precisely, for a given Borel set E in Rn the quantity P(t,E) =
0 u(x,t)dx measures the probability of finding the particle in E at time t.
5 E
Thegeneralproblemistofindu(x,t)whenthedynamicsofthesystemisknown
1 R
and some initial state is given.
:
v Regarding the dynamics of the system we shall deal with anomalous diffusions.
i
X More precisely with continuous time random walks (CTRW). For a comprehensive
introduction to the subject we refer to [6]. A CTRW in Rn is providedby a space-
r
a timeprobabilitydensityfunction,thekernel,J(x,t)definedinRn+1. Inthismodel
the particle has a probability density function u(x,t) of arrival at position x∈Rn
attime t>0whichdependsonthe eventsofarrivalatanyy ∈Rn,sometimesonly
on the events of arrival at any y in some neighborhood of x, at any previous time
s < t. Precisely, this dependence is given by the convolution in Rn+1 of J with u
itself. In other words, for t≥0 and x∈Rn
u(x,t)=(J ∗u)(x,t)= J(x−y,t−s)u(y,s)dyds. (1.1)
ZZRn+1
The physical condition of the dependence of the current position of the particle
only on the past (s<t) gives us the first natural condition on J,
(J1) suppJ ⊂Rn+1.
+
TheresearchwassupportedbyCONICET,ANPCyT(MINCyT)andUNL.
1
2
On the other hand, since J is a density in Rn+1, we must have
(J2)J ≥0, and
(J3)J ∈L1(Rn+1) and J(x,t)dxdt =1.
Rn+1
Following the notation iRnR[6], the density function defined in Rn by
λ(x)= J(x,t)dt
R
Z
is called the jump length probability function. Notice that from (J1), λ(x) =
J(x,t)dt. On the other hand, the waiting time probability function is given
R+
by
R
τ(t)= J(x,t)dx.
ZRn
Regarding the initial condition, let us first assume that the particle is localized
at the origin of Rn for t<0. In other words u(x,t)=δ (x) for t<0. Hence, since
0
u(x,t) for t≥0 needs to satisfy (1.1), from (J1) we must have that
u(x,0)= J(x−y,−s)u(y,s)dyds
ZZRn+1
= J(x−y,−s)u(y,s)dy ds
ZR−(cid:18)ZRn (cid:19)
= J(x−y,−s)δ (y)dy ds
0
ZR−(cid:18)ZRn (cid:19)
= J(x,−s)ds
ZR−
=λ(x).
In other words, the deterministic situation, the particle is at the origin for t < 0
produces immediately at time t = 0 a random situation modeled precisely by the
jump length probability function λ(x) associated to the density J.
Moregenerally,ifthepositionattimet<0oftheparticledistributesasindicates
the density f(x), then u(x,0) = (λ ∗f)(x). In this framework the basic initial
problemweareinterestedin,takesthefollowingform. GivenJ(x,t)andf(x),find
u(x,t) for (x,t)∈Rn+1 such that
+
u(x,t)=(J ∗u)(x,t), x∈Rn,t≥0;
(P) f(x), t<0;
 u(x,t)=
u(x,t), t≥0.

(cid:26)
Sometimes, to emphasize the data J andf in (P), we shallwrite P(J,f)for the

problem P and u(J,f) for its solution.
Letusobservethattheexpectedinitialconditionisattainedsince,takingt=0in
thefirstequationin(P) wegetu(x,0)=(J∗u)(x,0)= J(x−y,−s)f(y)dyds=
(λ∗f)(x).
RR
WeshallconsiderwidefamiliesofkernelsJ,butthereisone,theparabolicmean
value kernels, which plays a more significant role for our subsequent analysis. We
shall use H (for heat) to denote these special occurrences of J. Let us introduce
the most known of these kernels H (see [7] or [5]). Set W(x,t) to denote the
Weierstrass kernel for t > 0 and x ∈ Rn. Precisely W(x,t) = (4πt)−n/2e−|x|2/(4t).
Set E ={(x,t)∈Rn+1 :W(x,t)≥1} and H(x,t)= 1X (x,t)|x|2.
+ 4 E t2
3
As it is easy to check H satisfies properties (J1), (J2) and (J3) stated above.
Moreover, H satisfies also the following two properties
(J4) has compact support in Rn+1;
(J5) it is radial as a function of x∈Rn for each t.
The outstanding fact regarding H is given by its role in the mean value formula
for temperatures. If v(x,t) is a solution of the heat equation ∂v =△v in a domain
∂t
ΩinRn+1,then,for(x,t)∈Ωandrsmallenoughwehavethatv(x,t)= H (x−
r
y,t−s)v(y,s)dyds, where H denotes the parabolic r-mollifier of H. Precisely
r
RR
1 x t 1 |x|2
H (x,t)= H , = X (x,t) ,
r rn+2 r r2 rn E(r) t2
(cid:18) (cid:19)
with E(r) = {(x,t) ∈ Rn+1 : W(x,t) ≥ r−n}. The following figure depicts the
+
support E(r) of H .
r
x
t
Figure 1. Sets E(r) for n=1 and r = 1,1,1.
2 4 8
In the sequel, for any kernel J(x,t) and r > 0 we shall write J (x,t) to de-
r
note the parabolicapproximationto the identity givenby J (x,t)= 1 J x, t .
r rn+2 r r2
Moreover, the notation v (x,t) or even f (x) for functions depending on space
r r
(cid:0) (cid:1)
time or, only on the space variable will have always the same meaning. Precisely,
v (x,t)=r−n−2v(r−1x,r−2t) and f (x)=r−n−2f(r−1x).
r r
The results of this paper are in the spirit of those in [4] and [3]. Instead of
dealingwithgeneralizationofboundaryconditions,weareconcernedwithdiffusion
problems in the whole space Rn and the initial condition is generalized.
Letusstatethemainresultsofthispaper. Thefirstoneistheweakconvergence
to the heat equation.
Theorem 1. Assume that J(x,t) satisfies (J2), (J3), (J4) and (J5). Then, for
each ϕ in the Schwartz class of Rn+1, we have
1 ∂ϕ
lim (J −δ)∗ϕ=µ +ν△ϕ,
r→0r2 r ∂t
uniformly on Rn+1, where µ=− tJ(x,t)dxdt and ν = 1 |x|2J(x,t)dxdt.
2n
ThesecondresultconcernstheeRxRistenceofsolutionsforproRbRlem(P). Foragiven
Lipschitz function of order γ, f ∈ C0,γ(Rn), we denote by [f] the corresponding
γ
seminormof f. In the next statement C denotes the space of continuous functions.
4
Theorem 2. Assume that J(x,t) satisfies (J1), (J2), (J3) and (J4). Set α =
sup{β : J(y,s)dyds < 1}. Let f ∈ L∞(Rn) be given. Then there exists
s≤β
one and only one solution u(x,t) of (P) in the space (C ∩ L∞)(Rn+1). If f ∈
RR +
(L1 ∩L∞)(Rn), then u(x,t)dx = f(x)dx for every t ≥ 0. In particular,
Rn Rn
if f is a density function, so is u(·,t) for every t ≥ 0. Moreover, if f belongs to
(C0,γ ∩L∞)(Rn) we haRve that R
|u(x,t)−f(x)|≤C[f] (1.2)
γ
for (x,t)∈Rn×[0,α] and some C which does not depend on f.
Thenextresultwhichisinterestingbyitselfcontainsamaximumprinciplewhich
shallbeusedintheproofofTheorem4. Precisely,thesupremumoftheprobability
density functionin the future ofα=sup{β : J(y,s)dyds<1}coincides with
s≤β
its supremum in Rn×[0,α].
RR
Theorem 3. Let J be a kernel satisfying (J1), (J2), (J3), and (J4). Let w(x,t)
be a bounded function defined in Rn+1 such that
+
w(x,t)= J(x−y,t−s)w(y,s)dyds (1.3)
ZZ
for (x,t)∈Rn×[α,+∞). Then,
sup |w(x,t)| = sup |w(x,t)|.
(x,t)∈Rn+1 (x,t)∈Rn×[0,α]
+
Let us proceed to state the fourth result of the paper.
Theorem 4. For each H ∈ H there exists C > 0 such that for every r > 0 and
every f ∈(C0,γ ∩L∞)(Rn)
ku(H ,f)−uk ≤C[f] rγ,
r L∞(Rn+1) γ
+
where u is the temperature in Rn+1 given by u(x,t)=(W(·,t)∗f)(x).
+
Let us finally remark that in [2] the authors prove the Ho¨lder regularity for
solutions of the master equation associated to CTRW’s.
In Section 2 we prove the weak convergence of parabolic rescalings to a heat
equation. In Section3 we show existence ofsolutionfor the Cauchynonlocalprob-
lem. Section4is devotedtoprovethe maximumprinciple containedinTheorem3.
Finally, Section 5, deals with convergence of solutions of rescalings of (P) to tem-
peratures.
2. Some space time nonlocal parabolic operators and their weak
limit. Proof of Theorem 1
For 0<r <1, since J(y,s)dyds=1, applying Taylor’s formula we get
J (x−y,t−s)ϕR(Ry,s)dyds−ϕ(x,t)
r
ZZ
= J (x−y,t−s)(ϕ(y,s)−ϕ(x,t))dyds
r
ZZ
n
∂ϕ ∂ϕ
= J (x−y,t−s) (x,t)(y −x )+ (x,t)(s−t)
r ∂x i i ∂t
" i
ZZ i=1
X
5
1
+ (y−x,s−t)D2ϕ(x,t)(y−x,s−t)t+R(y−x,s−t) dyds,
2
(cid:21)
where D2 denotes the Hessian matrix of the second derivatives of ϕ with respect
to x and t and |R(x,t)|=O(|x|2+t2)23.
The last integral in the above identities can be written as the sums of the fol-
lowing seven terms,
n
∂ϕ
I = (x,t) (y −x )J (x−y,t−s)dyds ,
∂x i i r
i=1 i (cid:18)ZZ (cid:19)
X
∂ϕ
II = (x,t) (s−t)J (x−y,t−s)dyds ,
∂t r
(cid:18)ZZ (cid:19)
n ∂2ϕ 1
III = (x,t) (y −x )(y −x )J (x−y,t−s)dyds ,
∂x ∂x 2 i i j j r
ij=1,i6=j i j (cid:18) ZZ (cid:19)
X
n ∂2ϕ 1
IV = (x,t) (y −x )2J (x−y,t−s)dyds ,
∂x2 2 i i r
i=1 i (cid:18) ZZ (cid:19)
X
n ∂2ϕ 1
V = (x,t) (y −x )(s−t)J (x−y,t−s)dyds ,
∂x ∂t 2 i i r
i=1 i (cid:18) ZZ (cid:19)
X
∂2ϕ 1
VI = (x,t) (s−t)2J (x−y,t−s)dyds ,
∂t2 2 r
(cid:18) ZZ (cid:19)
and
VII = J (x−y,t−s)R(y−x,s−t)dyds.
r
ZZ
Since for t fixed J is radial as a function of x, then I, III and V vanish. For the
otherfourintegralsweperformtheparabolicchangeofvariables(z,ζ)=(x−y,t−s)
r r2
to obtain
∂ϕ
II = (x,t)r2 − ζJ(z,ζ)dzdζ
∂t
(cid:18) ZZ (cid:19)
n ∂2ϕ 1
IV = (x,t)r2 z2J(z,ζ)dzdζ
∂x2 2 i
i=1 i (cid:18) ZZ (cid:19)
X
∂2ϕ 1
VI = (x,t)r4 ζ2J(z,ζ)dzdζ
∂t2 2
(cid:18) ZZ (cid:19)
VII = J(z,ζ)R(rz,r2ζ)dzdζ.
ZZ
Finally, since, as a function of r close to zero, VI and VII are of order at least r3,
we see that
1 II IV ∂ϕ
lim [(J −δ)∗ϕ](x,t)= lim + =µ (x,t)+ν△ϕ(x,t),
r→0r2 r r→0 r2 r2 ∂t
(cid:18) (cid:19)
whereµandν aredefinedasinthestatementofTheorem1. Thatthusconvergence
is uniform in Rn+1 follows from the fact that ϕ is a Schwartz function and so VI
and VII converge to zero uniformly.
6
Lemma 5. For J =H we have that µ=−ν and the limit equation in Theorem 1
is the heat equation multiplied by a constant.
Proof. All we need to show is that
1
H(y,s)s dyds= H(y,s)|y|2 dyds. (2.1)
2n
ZZ ZZ
Let us compute both of them in terms of the Euler gamma function and the area
surface of the unit ball of Rn, Sn−1. On one hand we have that
1 |y|2
H(y,s)s dyds= X (−y,−s) s dyds
4 E(1) s2
ZZ ZZ
1 |y|2
=− dyds
4 s
ZZE(1)
1 0 |y|2
= dyds
4Z−41π ZB(cid:16)0,(2nsln(4π(−s)))21(cid:17) −s
1
1 0 1 (2nsln(4π(−s)))2
= ρn+1 dσdρds
4Z−41π −sZ0 ZSn−1
σ(Sn−1) 0 1 n+2
= (2nsln(4π(−s))) 2 ds
4(n+2)Z−41π −s
σ(Sn−1) 1 1 n n+2
= t(−ln(t)) 2 dt
4(n+2) t 2π
Z0 (cid:16) (cid:17)
= σ(Sn−1)nn+22 ∞e−θ(n+22)θn+22dθ
4(n+2)2n+22πn+22 Z0
= σ(Sn−1)nn+22 2 2n+22 ∞e−ζζn+22dζ
4(n+2)2n+22πn+22 (n+2)(n+2)n+22 Z0
σ(Sn−1)nn+22 n+4
= Γ .
2(n+2)n+26πn+22 (cid:18) 2 (cid:19)
On the other,
1 1 |y|2
H(y,s)|y|2 dyds= X (−y,−s) |y|2 dyds
2n 8n E(1) s2
ZZ ZZ
1 |y|4
= dyds
8n s2
ZZE(1)
1 0 |y|4
= dyds
8nZ−41π ZB(cid:16)0,(2nsln(4π(−s)))21(cid:17) s2
1
1 0 1 (2nsln(4π(−s)))2
= ρn+3 dσdρds
8nZ−41π s2 Z0 ZSn−1
σ(Sn−1) 0 1 n+4
= (2nsln(4π(−s))) 2 ds
8n(n+4)Z−41π s2
σ(Sn−1) nn+244π 1 1 n+2
= (t(−ln(t))) 2 dt
8(n+4)2n+24πn+24 Z0 t2
7
= σ(Sn−1)nn+224π ∞e−θ(n+22)θn+24dθ
8(n+4)2n+24πn+24 Z0
= σ(Sn−1)nn+224π 2 2n+24 ∞e−ζζn+24dζ
8(n+4)2n+24πn+24 (n+2)(n+2)n+24 Z0
σ(Sn−1)nn+22 n+6
= Γ .
(n+4)(n+2)n+26πn+22 (cid:18) 2 (cid:19)
Now, since Γ(z+1)=zΓ(z), we have that 1 Γ n+6 = 1Γ n+4 and the proof
n+4 2 2 2
is complete. (cid:3)
(cid:0) (cid:1) (cid:0) (cid:1)
3. Existence of solutions for (P). Proof of Theorem 2
LetJ(x,t)beakerneldefinedinspacetimeRn+1 satisfying(J1), (J2),(J3) and
(J4). Letf ∈L∞(Rn)begiven. Followingtheideasin[4],[3]and[1]weshallsolve
(P) byiteratedapplicationoftheBanachfixedpointtheorem. From(J3) and(J4),
α= sup{β : J(x,s)dxds < 1} is positive and finite. For the first step in the
s<β
use of the fixed point theorem in the Banach space B = (C ∩L∞)(Rn ×[0,α])
RR 1 2
with the L∞ norm.
As in the statement of (P) set
f(x), t<0;
v(x,t)=
v(x,t), t∈[0,α],
(cid:26) 2
wherev ∈B . Sincev isboundedonRn×(−∞,α]andJ ∈L1(Rn+1),theintegral
1 2
g(x,t):= J(x−y,t−s)v(y,s)dyds
ZZRn×(−∞,α2]
is absolutely convergentfor (x,t)∈Rn×[0,α]. Let us prove that, as a function of
2
(x,t) ∈ Rn×[0,α] the function g belongs to B . In fact, from the definition of g
2 1
we see that
|g(x,t)|≤ Jdyds kvk ≤sup{kfk ,kvk }.
∞ ∞ ∞
(cid:18)ZZ (cid:19)
Let us check the continuity of g. For h ∈Rn and k ∈R such that (x+h,t+k)∈
(−∞,α], we have that
2
|g(x+h,t+k)−g(x,t)|≤ |J(x+h−y,t+k−s)−J(x−y,t−s)||v(y,s)|dyds
ZZ
≤ω ( |h|2+k2)kvk ,
1 ∞
q
where ω is the modulus of continuity in L1 of J. Hence for v ∈ B we also have
1 1
that g ∈B when restricted to the strip Rn×[0,α].
1 2
Define T :B →B by T v =g.
1 1 1 1
Let us now prove that T is a contractive mapping in B . Let v and w be two
1 1
functions in B . Let (x,t)∈Rn×[0,α]. Then, with
1 2
f(x), t<0;
w(x,t)=
w(x,t), t∈[0,α],
(cid:26) 2
8
we have that
T v(x,t)−T w(x,t)= J(x−y,t−s)(v(y,s)−w(y,s))dyds
1 1
α
ZZs≤2
= J(x−y,t−s)(v(y,s)−w(y,s))dyds.
α
ZZ0<s≤2
Hence
kT v−T wk ≤ sup J(x−y,t−s)dyds kv−wk
1 1 ∞ (x,t)∈Rn×[0,α2]ZZ0<s≤α2  ∞
 
Now, from the definition of α, and (J1)
J(x−y,t−s)dyds= J(z,σ)dzdσ
α α
ZZ0<s≤2 ZZt−2<σ≤t
= J(z,σ)dzdσ
ZZ0<σ≤t
≤ J(z,σ)dzdσ =:τ <1.
α
ZZ0<σ≤2
So that kT v−T wk ≤ τkv−wk . Hence T is a contractive mapping in B .
1 1 ∞ ∞ 1 1
So that there exists a fixed point u ∈B for T ; T u =u . In other words
1 1 1 1 1 1
u (x,t)= J(x−y,t−s)u (y,s)dyds (3.1)
1 1
ZZ
for x∈Rn and 0≤t≤ α.
2
Let us check that
u (x,t)dx= f(x)dx
1
ZRn ZRn
for every 0 ≤ t ≤ α, when f ∈ L1(Rn). Since u can be realized as the limit
2 1
of the sequence of iterations of T applied to any function v ∈ B , we may take
1 1
v(x,t)=f(x) as the startingpoint. In doingso we see thatthe integralin variable
x of |Tmf(x,t)| is less and equal to |f|dx. In fact, from (J3), we see
1
R
|T f(x,t)|dx= J(x−y,t−s)f(y)dyds dx
1
Z Z (cid:12)ZZ (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
≤ (cid:12) J(x−y,t−s)dxds |f(cid:12)(y)|dy
Z (cid:18)ZZ (cid:19)
= |f|dy.
Z
Hence, inductively, assuming |Tmf(x,t)|dx≤ |f|dx we have
1
Tm+1f(x,t) dx= |RT (Tmf)(x,t)|dx R
1 1 1
Z Z
(cid:12) (cid:12)
(cid:12) (cid:12) = J(x−y,t−s)Tmf(y,s)dyds dx
1
Z (cid:12)ZZ (cid:12)
(cid:12) (cid:12)
= (cid:12)(cid:12) J(y,t−s)T1mf(x−y,s)dyds(cid:12)(cid:12)dx
Z (cid:12)ZZ (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
9
= J(y,t−s)f(x−y)dyds+
Z (cid:12)ZZs<0
(cid:12)
(cid:12)(cid:12) + J(y,t−s)T1mf(x−y,s)dyds dx
ZZs>0 (cid:12)
(cid:12)
(cid:12)
≤ J(y,t−s) f(x−y)dx dyds+ (cid:12)
ZZs<0 (cid:12)Z (cid:12)
(cid:12) (cid:12)
+ J((cid:12)(cid:12)y,t−s) T1m(cid:12)(cid:12)f(x−y,s)dx dyds
ZZs>0 (cid:12)Z (cid:12)
(cid:12) (cid:12)
≤ |f|dx. (cid:12) (cid:12)
(cid:12) (cid:12)
Z
With the same arguments we can conclude that Tm+1f(x,t)dx = f(x)dx for
1
0≤t≤ α. The resultthenfollowssince,forf ∈L1∩L∞,wehavethatTmf tends
2 R R 1
to u also in C([0,α],L1(Rn)). In fact, if we prove that
1 2
|||Tm+1f −Tmf|||≤τm|||T1f −f||| (3.2)
1 1 1
where |||v||| = supt∈[0,α]kv(·,t)kL1(Rn), then T1mf is also a Cauchy sequence in
2
C([0,α],L1(Rn)). Since Tmf converges uniformly to u we get the desired preser-
2 1 1
vation of the integral. Let us prove (3.2).
Let us first check that Tmf is continuous as a function of t∈[0,α] with values
1 2
in L1(Rn) for each m. Take t and t+h two points in [0,α]. Then
2
|Tmf(x,t)−Tmf(x,t+h)|dx
1 1
Z
= J(x−y,t−s)Tm−1f(y,s)dyds
1
Z (cid:12)ZZ
(cid:12)
(cid:12)(cid:12) − J(x−y,t+h−s)T1m−1f(y,s)dyds dx
ZZ (cid:12)
(cid:12)
= hJ(z,t−s)−J(z,t+h−s)iT1m−1f(x−(cid:12)(cid:12)z,s)dzds dx
Z (cid:12)ZZ (cid:12)
(cid:12) (cid:12)
≤ (cid:12)(cid:12) |J(z,t−s)−J(z,t+h−s)| T1m−1f(x−z,s) d(cid:12)(cid:12)x dzds
Z Z (cid:18)Z (cid:12) (cid:12) (cid:19)
(cid:12) (cid:12)
≤ |f(x)|dx |J(z,t−s)−J(z,t+(cid:12)h−s)|dzds, (cid:12)
Z Z Z
which tends to zero when h→0 because J ∈L1(Rn+1).
Similar calculations show that Tmf is a Cauchy sequence in the |||·|||. In fact,
1
for t∈[0,α] we have
2
Tm+1f(x,t)−Tmf(x,t) dx
1 1
Z
(cid:12) (cid:12)
=(cid:12) J(x−y,t−s) T(cid:12)mf(y,s)−Tm−1f(y,s) dyds dx
1 1
Z (cid:12)(cid:12)ZZ (cid:16) (cid:17) (cid:12)(cid:12)
= (cid:12)(cid:12) J(x−y,t−s) T1mf(y,s)−T1m−1f(y,s) (cid:12)(cid:12)dyds dx
Z (cid:12)Z Z0≤s≤t (cid:12)
(cid:12) (cid:0) (cid:1) (cid:12)
≤ (cid:12)(cid:12) J(z,t−s) Tmf(x−z,s)−Tm−1f(x−z,s) d(cid:12)(cid:12)x dzds
1 1
Z Z0≤s≤t (cid:18)Z (cid:19)
(cid:12) (cid:12)
(cid:12) (cid:12)
10
= J(z,t−s) Tmf(x,s)−Tm−1f(x,s) dx dzds
1 1
Z Z0≤s≤t (cid:18)Z (cid:19)
(cid:12) (cid:12)
(cid:12) (cid:12)
≤ sup Tmf(x,s)−Tm−1f(x,s) dx J(z,t−s)dzds
s∈[0,α2]Z (cid:12) 1 1 (cid:12) ZZ0≤s≤α2
=τ|||Tmf −T(cid:12)m−1f|||, (cid:12) 
1 1
hence
|||Tm+1f −Tmf|||≤τ|||Tmf −Tm−1f|||.
1 1 1 1
By iteration we obtain (3.2).
Letusobservethatsinceu (x,t)canbeobtainedastheiterationofT startingat
1 1
anyfunctionv inB ,wecaninparticulartakev astheconstantfunction s(f)−i(f),
1 2
where s(f) = supf and i(f) = inff. Then v = vX{0≤t≤α} +fX{t<0}, so that
2
i(f) ≤ v ≤ s(f) everywhere. From (J2) and (J3) we also have i(f) ≤ T v ≤ s(f)
1
on Rn×[0,α]. The same argument shows that for every iteration Tkv of T v we
2 1 1
have i(f)≤Tkv ≤s(f). Since u is the uniform limit of Tkv we get
1 1 1
i(f)≤u (x,t)≤s(f)
1
onthestripRn×[0,α]. Sofarwehaveexistenceandmasspreservationfort∈[0,α].
2 2
NowproceedinductivelybycoveringR+ withintervalsofthetype[(i−1)α,iα].
2 2
The first step, i = 1 is precisely the one described above. Assume that u ∈B =
i i
(C∩L∞)(Rn×[(i−1)α,iα]) for i=1,...,j have been built in such a way that
2 2
u (x,t)= J(x−y,t−s)u (y,s)dyds,
i i
ZZ
with
u (x,t), t<(i−1)α;
u (x,t)= i−1 2
i u (x,t), (i−1)α ≤t≤iα.
(cid:26) i 2 2
Moreover, u (x,t)dx= f(x)dx for (i−1)α ≤t≤iα,
Rn i Rn 2 2
R R i(f)≤ui(x,t)≤s(f) (3.3)
for every (x,t) ∈ Rn ×[(i−1)α,iα], and u (x,(i−1)α) = u (x,(i−1)α) for
2 2 i 2 i−1 2
every x.
DefineB asthespace(C∩L∞)(Rn×[jα,(j+1)α])withthecompletemetric
j+1 2 2
induced by the L∞ norm. For v ∈B , define
j+1
T v(x,t)= J(x−y,t−s)v(y,s)dyds
j+1
ZZ
with
u (x,t), t<jα;
v(x,t)= j 2
v(x,t), jα ≤t≤(j+1)α.
(cid:26) 2 2
As in the case of i = 1, it easy to check that with (x,t) ∈ Rn ×[jα,(j +1)α],
2 2
T v ∈ B . Hence T : B → B . It is also easy to prove that T is
j+1 j+1 j+1 j+1 j+1 j+1
contractive on B with the same rate of contraction τ obtained when i=1.
j+1
Also, with the same argumentas inthe casei=1,with u (x,t)dx= f(x)dx
j
when t≤jα we have that for t∈[jα,(j+1)α]
2 2 2 R R
u (x,t)dx= f(x)dx.
j+1
ZRn ZR