A PROBABILISTIC APPROACH TO THE TRACE AT THE BOUNDARY FOR SOLUTIONS OF A SEMILINEAR PARABOLIC PARTIAL DIFFERENTIAL EQUATION

We use the path-vMued process called the "Brownian snake" to investigate the trace at the boundary of nonnegative solutions of a semilinear parabolic partial differential equation. In particular, we characterize possible traces and in dimension one we prove that nonnegative solutions are in one-to-one correspondence with their traces at the origin. We also provide probabilistic representations for various classes of solutions. This article is dedicated to the memory of Roland L. Dobrushin.


Introduction and Statement of the Results
The goal of this work is to develop a probabilistic approach for studying the trace at the boundary of positive solutions to the semilinear parabolic equation Ou This approach has been inspired by our previous work [14] and the recent paper of Dynkin and Kuznetsov [7], which both dealt with the trace at the boundary for related semilinear elliptic par- tial differential equations.Our main probabilistic tool is the path-valued process called the Browniau snake, whose connections with semilinear partial differential equations have been investi- gated in several recent papers [10,13,14].Since the Brownian snake is closely related to the super-Brownian motion, part of these connections can be viewed as a reformulation of Dynkin's   important work on the relation between superprocesses and partial differential equations [3][4][5].
However, we think that the Brownian snake is more tractable, although less general, for certain applications.In particular, it is not clear how to derive the results of [14] or of Section 4 of the present work, using only the theory of superprocesses.On the other hand, it is very plausible that superprocesses can be applied to extend a significant part of the present work to more general equations where the nonlinear term u 2 is replaced by u s for 1 < c _< 2.
The problem of the trace at the boundary for semilinear elliptic or parabolic equations of type (1) has also been studied recently by analytic methods.See in particular, Marcus and Vron [15].
Very recently, as the final version of this work was in preparation, we received the note [16], which announces results that generalize some of the statements below (Theorem 1 and the analytic part of Theorem 4) to equations of form (1) with a nonlinearity u p instead of u2, for any p > 1.Nonetheless, we feel that it is worth developing the probabilistic approach, which in the case p 2 gives slightly more precise results (in contrast to [16], we are able to characterize all possible traces; see Proposition 2 below) and also yields explicit probabilistic formulas for the solutions.
Let us now state our main results.For y E d and r > 0, we denote by B(y, r) the open ball of radius r centered at y.
Theorem 1: Let be a domain in d and let u C1,2((0,cx)x ) be a nonnegative solution of (1) in (O, oc) x a. Set {y e f;Vr > 0 lim / u(t z)dz There exists a Radon measure , on f\A such that, for every 9 G Cc(a\A), (u,)-lim / u(t z)(z)dz.

\A
The pair (A,u) is called the trace of u.
The method of proof gives precise information about the behavior of u near (0, y), when y f\A.See Lemma 6 and the remark following the proof of Theorem 1 in Section 3.
Our second result gives a characterization of possible traces of a solution.It is obviously analogous to Theorem 1.3 of Dynkin and Kuznetsov [7].In this statement, "polar" means "polar with respect to Brownian motion in Rd., Proposition 2: Let A be a closed subset of and let u be a Radon measure on \A.In vai (A..) of a no.n a iv o1.ion of in ary and sufficient that the following conditions hold: (a) u does not charge polar sets.(b) Let Ex(u) {y G A, Vr > O,u(B(y,r)) cx} be the set of explosion points of u and let H be a polar subset of a such that A\H is closed and H V Ex(u)= .Then H O.
Proposition 2 resolves the problem of the existence of a solution with a given trace.We now address the uniqueness problem when f-Nd.We use the probabilistic representation of solutions involving the Brownian snake (Ws, s >_ 0) under its excursion measures Nx, x E Nd.A brief presentation of these probabilistic objects is given in Section 2 below.Here, we simply note that, under Nx, for every s >_ 0, Ws:[0,s]--d is a finite path in [d started at x and with "lifetime" s" The lifetime process (s,s _> 0) is distributed under N x according to the It measure of positive excursions of a linear Brownian motion.For every _> 0, we set t-{Ws(t),sO,st}, which corresponds to the values at time t of all paths W s with lifetime s >-t.Let u be a finite measure on Nd not charging polar sets.For every t > 0, we can consider the finite measure on paths with lifetime t defined by #z, t(dw) J u(dy)Pt(X, Y)Ptu(dw), where pt(x,y) is the Brownian transition density, and Ptu denotes the law of the Brownian bridge from x to y in time t.Following [2], we can associate with the measure #t x an additive functional of the Brownian snake with initial point x.We denote by A t'u-1Ats'U,s _> 0) this additive functional.By a monotonicity argument, the definition of A t'u can be extended to the case when u is a countable sum of finite measures not charging polar sets.
Proposition 3: Suppose that f-d and let (A,) be a pair satisfying the conditions of Pro- position 2. Then there exists a maximal nonnegative solution of (1) with trace (A,), and it is given by the probabilistic formula u(t,x)-2Nx(YtAeq))+2N x l{ftA=O}(1-exp(-A 2)) (2) If A is polar, u is the unique nonnegative solution with trace (A,u).On the other hand, if d >_ 2, and if (A,0) is a possible trace, with A O, then there exist infinitely many solutions with the trace (A, 0).For a general choice of (A,u) in dimension of d _> 2, it is not easy to decide whether there is a unique associated solution.A natural guess would be that uniqueness holds if and only if A-Ex(u), but we were unable to prove this.
In dimension d 1, uniqueness always holds.This follows from the next theorem, where the probabilistic representation of solutions is made somewhat more explicit by the following considerations.For every t > 0, denote by Lt= Lts, s >_ 0) the local time process of (s,s >_ 0) at level t.Then let X be the random measure on N" supported on S defined by (Xt, 9) J dnts(Ws(t))" 0 Then [9] the distribution of (Xt, t > 0) under Nx is the canonical measure of super-Brownian motion with initial point x.When d 1, it follows from well-known results (see e.g., Sugitani  [18], Theorem 1) that the measures X are absolutely continuous, and, more precisely, we may write Xt(dY Yt(Y)dy where the process (Yt(y);t > 0, y ) is jointly continuous.With this notation at hand, we can state our last result, which is analogous to the main result of [14].
Theorem 4: Let d-1 and f-R.The formula u(t,x)-2Nx(tC?A O)+2N x (l{tA:O}(1-exP-1/2]u(dY)Yt(y))) (3)   gives a one-to-one correspondence between nonnegative solutions of (1) and pairs (A,p), where A is a closed subset of and, is a Radon measure on NA. gu is given by formula (3), the trace of u is (A, The paper is organized as follows.In Section 2, we recall the basic facts about the Brownian snake, its additive functionals and the connections with solutions of (1).Several results of this sec- tion are valid in a much greater generality, but we limited ourselves to those facts that are needed in the proofs of the following sections.In Section 3, we give the proof of Theorem 1 and Pro- positions 2 and 3. Several arguments of this section are inspired from [7], although our definition of the trace is different and more analytic in the spirit of [15] and [16].Finally, in Section 4, we prove Theorem 4 following ideas from [14].

Analytic Preliminaries
In view of the probabilistic representation, it will be preferable to deal with a slightly modified form of equation ( 1).We shall be interested in solutions of the equation 0__u_u + 1/2An 2u (4) 0t in oc, 0) x f.Up to a trivial scale parameter, the change of variables t t reduces the study of (1) to that of (4).Rather than proving the results stated in the introduction, we shall prove below the equivalent statements concerning (4).As we shall deal only with nonnegative solu- tions, the word "solution" will always mean "nonnegative solution".A solution of (4) in the do- main D of N x d is thus a nonnegative function u E C1'2(D) such that (4) holds pointwise in D.
Brownian motion in d will be denoted by .For every (t,x) x Nd, we will denote by Pt, x the probability measure under which starts at x at time t (r is thus defined for r > Let D be a bounded domain in NxNd.We denote by 7 the exit time of from D'r- inf{r; (r, r) D}.A measurable subset H of OD is called total if Pt, x((r, r) H) 1, for every (t,x) D. The following maximum principle for solutions of (4) can be found, in a more general form, in the appendix of Dynkin [4].
Maximum principle: Let u 1 and u 2 be two (nonnegative) solutions of (4) in D. Assume that u 1-u2 is bounded above and that there exists a total subset H of OD such that for every (r, y) H, limsup (Ul(t,x) -u2(t,x)) <_ O. Ds(t,x)-+(r,y) Then u I _< u 2.

The Brownian snake
It will be convenient to work in a time-inhomogeneous setting as described in [13], Section 2.1.Therefore, we slightly extend the notation of the introduction as follows.For every fixed (t,y) N x Nd, we denote by rt, x the space of all finite continuous paths w'[t,]N d such that w(t)-x (here w can be any real number in [t, c)).We denote by (Ws, s >_ 0) the Brownian snake in t,x and by Nt, x its excursion measure away from the trivial path in t,x with -t (see [13], Section 2.1).Under Nt, x, each W s:[t,s]--Nd is a finite path in Nd, started from x at time t and stopped at time s" The distribution of (s,s >_ 0) under Nt, is the It8 measure of excursions of linear Brownian motion above level t.Informally, the path W s extends itself by adding little pieces of Brownian motion when s increases, and erases itself when s decreases.
For w gt, x, we also denote by P*w the law of the Brownian snake started at w and stopped when it first hits the trivial path of Irt, x (equivalently when first hits t).
The range and graph of the Brownian snake are defined respectively as > 0, t _< _< > 0, t _< _< Note that 1 and are compact connected subsets of [R d and R x d, respectively.It is easy to prove that for every g > 0, Nt, x(J n B(x g)c 5 O) Cd e 2 Nt, x(On[t+e,) xd 0)--(2e) -1 where c d is a positive constant (see e.g., [10]).
For t' > t, we also set Jt[t,t, {Ws(r); s >_ O,t <_ r <_ ' A s}" Let r > 0 and r() Nt, (%[t, + ] N B(x, r) c =/: ).We shall frequently use the fact that limCr()-0. (5) One way to derive (5) is to check from the Kolmogorov lemma that the paths Ws, s > 0, satisfy a uniform HSlder condition in the variable t.In fact, results much more precise than (5) are known.See in particular, Theorem 3.3 of [1].
As in Section 1, we write fr-{Ws(r)'s>-0,s>-r}, for every r_>t.Note that Jr} x 5r-n ({r} x Nd).By a result of Dynkin [4], if H is a Borel subset of N d and r > t, Nt, x(f r N H q)) is positive if and only if H is not polar.This fact explains the relevance of polar sets in our main results.

Exit measures
We now consider a domain D in x d of the type (-c,a)x Q.As previously, we denote by r the first exit time of Brownian motion from D and we also write r(w) for the first exit time from D of a finite path w.
Assuming that (t,x)E D, we can define the exit measure of the Brownian snake from D under the excursion measure N x (see [13], Section 2.2 or [10]).This measure, denoted by XD, is a finite measure supported on (r(W,), W,(r(W))); s > 0, r(W) < c} C 0D.In the special case D D a -cx3, a) x d, we can write X Da 5a (R) Xa, where X a is a finite measure on d.This definition is consistent with the notation of Section 1.
Let g be a nonnegative measurable function on 0D, which is bounded on the sets 0D A ([r,c)xEd).By applying Theorem 4.2 of [10] to the special case when the underlying spatial motion is a space-time Brownian motion, we get that the function u(t,x) -Nt, x(1-exp-<XD, g>), (t,x) ED, solves the integral equation r .(t,+ From this integral equation and the classical connections between (space-time) Brownian motion and the heat equation, we easily deduce that the function u solves (4) in D.
Conversely, solutions of (4) can be represented in the previous form.
Mean value property: Let u be a solution of (4) in D. Let D' be a subdomain of D of the form D'--(c, a') ', where a' < a and ' is a relatively compact subdomain of .Then, for every (t, x) D', u(t, x) Nt, x(1 exp <X D', u)).
This follows from the maximum principle applied to u and to the function Nt,(1--exp (xD', u)) in the domain (r,a')xf',r < a'.We have seen that v solves v(t, x) (4) in D' and from the integral equation satisfied by v, it is easy to verify that v has boundary value u on a total subset of 0D'.
In the special case -Nd, we can take '-B(0, R) and then let R--<x to obtain u(t,x) Nt, z(1 -exp-(Xr, u(r )}) for t < r < a.
Remark: The previous mean value property can be stated in a much more general form.

Additive functionals
We now take D (-oe, 0)x f for simplicity.In order to construct more general solutions of (4), we introduce additive functionals of the Brownian snake.Let be a finite measure on f.We first assume that has a finite energy in the classical potential-theoretic sense: / /(dY)(dZ)fd('Y-Z')<cx3' where fl(r)-l,f2(r)-log+(1/r), and fd(r)--r 2-d ifd>3.If t<0 and x,y, we denote tO by P xy the law of the Brownian bridge started from x at time t and conditioned to be at y at time 0. We can view Ptx as a probability measure on the set {w e qirt, x, 0}.We also denote by ptO, the law of the bridge conditioned to remain inside .We set xy f #t where pr(X,y) stands for the transition density of Brownian motion killed when it exits Ft.We can interpret t,x as the law of the h-transform of a space-time Brownian motion in D started at (t,x) corresponding to the harmonic function h(t,x)-f u(dy)p t(x, y).
The energy of t,x with respect to the Brownian snake in t,x is easily computed from [12]   Proposition 1.1" e("t, x) 2Et, z drh(r, r) 2 2 dr dyp_ t(x, y) u(dz)p r(Y, z) Thus, ("t, x)2 f / u(dz)u(dz')Ft, x(z'z')' 0 where Ft, (z, z') 2 f dr f dypr t(x, y)p r(Y, z)p r(Y, Z').Elementary estimates give the d bound Ft,(z,z' C(t)fd(lZ-z'l), with a constant C(t)< depending only on t.Therefore our assumption on guarantees that (Pt, x) < " We can then use [2] Theorem 5 to construct the continuous additive functional associated with the measure t,z, which we denote by A '-(A ', s _> 0) (we drop (t,x)in the notation since it is understood that we work under the measure Nt, x)" For any nonnegative measurable function F on t,x, we have 0 An approximation for A u' can also be given as follows.
or every r >0, set D , r) x a, and make the convention that h 0 on , 0) x 0a.Then, A a lim(xDr, h} (7) r0 in L2(Nt, x) (compare with [13], Section 4.4).This follows from standard energy calculations" for t < r < O, (xDr, h)-A 'r, where A u''r is the additive functional whose associated measure t,x,r is the image of t,x under the mapping ww][t,r].Then observe that Pt, x,r converges to in the energy norm.

t,x
It is easy to extend the previous construction to the case when u is a finite measure that does not charge polar sets.By a classical result, we may write u-limUn, where the measures u n have a finite energy.Then, we may define A'-limT Aun' for every s0, Nt, x a.e.By (6), Nt, z(AU) h(t,x)< , so that A< , Nt, x a.e.In addition, from the time-reversal invariance property of additive functions ([2] Theorem 5), we have also A-A'= -A s ), for every s O, Nt, x a.e.It follows that the convergence of Ash' to A ' is uniform in s 0, and we easily conclude that A u' is also a continuous additive functional for which formula (6) still holds.
Proposition 5: Let be a finite measure on that does not charge polar sets.
The function uu, is the unique nonnegative solution of the integral equation For every In particular, uu,t solves (4) in (-c,0).In addition, for any bounded continuous function on , we have lim J (t, x) / (dx)(x).
Proof: Write u-u, and h(t,x)-f(dy)pt(x,y as previously.We use the strong Markov property under t,x [10] and the fct that A "' is an additive functional to obtain u(t,x) Nt, dA' aexp (A a A'a) Nt, x dA'aE(exp-A a) 0 0 By (6) and the form of the measure #t,, we know that dA 'fl a.e.W s is a finite path in Q started at time t and stopped at time 0. We can use [10] Proposition 2.5 to compute Ews* (exp A).Approximation (7) allows us to verify that, under es,A-E A(wi), in the notation of this proposition, and so we get t,z a.e., dA ' a.e., 0 Hence, 0 by (6).The proof of ( 8) is then completed by routine calculations.Using simple properties of the law of the Brownian bridge, we get u(, z) h(, ) 2 f u(d)p a_ t(, )E a (r, w(r))exp (s, w(s)) e(, h(, (e), e( )p (,   h(t, x) 2Et, x dr u(r, r) 2 where in the last equality we used the previous displayed formula for u.
Since h is space-time harmonic (for + A)in D, it is easy to verify that u solves (4).The uniqueness of the nonnegative solution of (6) follows by the same arguments as in the uniqueness part of the proof of Proposition 4.1 in [13].Alternatively, this uniqueness can also be obtained as very special case of Theorem 1.1 in [6].
To prove the last assertion, we first assume that has a finite energy.Consider the case 1, and note that u(t,x) h(t,x) f (dy)p_ t(x, y). which tends to 0 as tO by the dominated convergence theorem.Hence, / dxu(', x)-lim / dxh(,, x)-(u 1).lim tTo tTo Then, it is sufficient to consider the case 0 _ p _ 1.In this case, lim sup f dxa(x)u(t, x) <_ lim sup f dx(x)h(t, x) (,, ), tT0 J tT0 J and replacing p by 1yields the desired result.FinMly straightforward monotonicity argu- ment makes it possible to drop the finite energy assumption.
Remark: Dynkin and Kuznetsov [6] solve integrM equations of a more generM form than (6) in terms of additive functionals of superprocesses.As consequence of their results ([6], Theorem 1.3), the condition that u does not charge polar sets is necessary for the existence of (nonnega- tive) solution of (6).A proof of this fact using the Brownian snake can also be given Mong the lines of the proof of Proposition 4.3 in [13].

Singular solutions
It is also easy to get a probabilistic representation for solutions that tend to infinity on a part of the boundary.We limit ourselves to a special case that will be needed later.Let D (-cx3, 0) as previously.Define under Nt, x the graph of the Brownian snake in D by {w(); _> 0, _< _< (w) .
Note that the support of X D is contained in Df3 0D.Let U be a measurable subset of cOD, and let u be a finite measure on d not charging polar sets.Write A u-A u' for simplicity.The function u(t, x) Nt, x(O D rq U # O) + Nt, x( D U 0;1 exp A), (t, x) D, is a solution of (4) in D. The easiest way to verify this fact is to use the special Markov property for the nrownian snake [13].
Note that u(t,x)-Nt, x(1-e-Z), where_ Z-A+ oo-1 oD Let D'-(-oo, a)x ft' be a subdomain of D, with a < 0 and ' C .Then if { (wi, I) denote the excursions of the nrownian snake outside D' (see [13]), it is easy to verify that Z Z(wi), Nt, x a.e. for (t,x) D (use (7) to deal with the additive functional part).The special Markov property then gives u(t,x) Nt, x(1 -exp--{xD',u)) and, as we saw previously, this implies that u solves (4) in D'.
Let (r, y) belong to the relative interior of U in OD.Suppose that either r 0 or y is regular for c (with respect to Brownian motion).Then, by writing u(t,x)>_ Nt, x(1-exp-n(xD, 1u)), it is easy to verify that u has boundary value + oc at (r, y).

The Trace of a Solution
The next lemma is the key to the proof of Theorem 1. tion 5.
Recall the notation uu, ft from Proposi- not charging polar sets and such that lim]tT0 dyu(t,y)p(y)-/u(dy)p(y), n\A for every E Cc(\A).Theorem 1 follows.

El
Remark: The pair (A,u) is called the trace of .As a consequence of Lemma 6, we also arrive at the following fact that will be needed alter.Let u I and u 2 be two solutions of (4) in OO, 0) X 1 and oo, 0) x 22, respectively, with respective traces (A1,/21) and (A2, 2)" Sup- pose that there exists a ball B such that B C (fl\A1)r3 (f2\A2) and the measures t/1 and u2 co- incide on a neighborhood of B. Then, lim sup tt l(t, X) u2(t x) O. tT0 This follows from_ (9) by comparing u I and u 2 with the solution uu, B,, where B' is a suitable neighborhood of B and u stands for the restriction of Ul, or u2, on B'.
The proof of Proposition 2 depends on the following two lemmas.
Lemma 7: Let (A,u) be the trace of a solution u of (4) in (-(x,O)xf.Property (b) of Proposition 2 holds for the pair (A, u).
Proof: Let H be as in property (b), and suppose that He0.Then, let xGH and let B be an open ball centered at x such that B N A is polar and u(B') < (x for a certain neighborhood B' of B. The existence of follows from our assumptions on H. Denote by u B, the restriction of u on B' and write AuB''-.Finally, let (9 71A) denote the open e-neighborhood of 71A in N d and for (t, y) G D" c, O) x B, set ve(t,y Nt, u({%[t,olnBC :/: 0} U {Yo n (B nA) : +Nt, u({%[t,o]nB-O}n{2on(B nA)-O}; 1-exp-A ).
By combining the previous observations, we have limsup (u(t, y) re(t, y)) <_ O, (t,y)--,(r,z) for every (r,z) 0D.In particular, u-v is bounded above on the sets of the form (a,O) B, and the maximum principle implies that u _< v in D. By letting 0, we get tl(t,y) _ Nt, y({'-J[t,O] r-'l B c : 0} U {o r-"l (B r'l A) -)/: 0}) 1-exp-A2B' 1 exp A2B'), because the polar set B N A is not hit by 0, Nt, v a.e.This bound and (5) imply that u B limsup (u(t,y)-Nt, y(1-exp-Aoo ))_<0, (t, )(0, ) uniformly when z varies over a compact subset of B. Clearly, this implies that B N A-0 which gives a contradiction.
Lemma 8: Let A be a closed subset of N d and let v(t,y)-Nt, y(fo fl A 0).The set H-{y tto s polar.
B(, Proof: It is relatively easy to give an analytic proof along the lines of the proof of Lemma 5.1 in [7].We will give a probabilistic argument that does not use the connections with partial differ- ential equations.It is sufficient to prove that, if B is an open ball such that, for a certain sequence tnTO, sup I v(tn, z)dz < oo, J then B r3 A is polar.

B
First note that, by the first-moment formula for Xt, if t < t n and x E [d, Nt, x((X, n, 1Bv(tn, ))) Et, x(1B(tn)v(tn,tn)), and so by our assumption, 1BV(tn, ))) < sup Nt, x((Xtn n: n < On the other hand, by the special Markov property [13], Nt, x((Xtn, lBv(tn,.)))is the expected number (under Nt.x) of the "excursions" of the Brownian snake in [tn, oo x N d that start from {tn} x B and hit {0} x A. Letting n+oo, we get from the previous bound that Nt, x(Card{s >_ 0; (s O, Ws(O B f3 A)) < oo.
This implies that the set {w;(-0, w(0) B r3A} is semipolar for the Brownian snake, hence also Mt, x-polar in the terminology of [9] (because the Brownian snake is a symmetric Markov process, see [9]).By the Dynkin result recalled at the end of Subsection 2.2, this is equivalent to saying that B V1 A is polar.
Proof of Proposition 2: The necessity of condition (a) has already been established in the construction of the trace, and the necessity of (b) follows from Lemma 7. To prove the sumciency of (a) and (b), we may clearly take f-Nd (otherwise we construct a solution in (-oo, 0)x Nd with trace (A tO fY, u)and then we restrict it to (-oo, 0)x f).
Let (Kn) be an increasing sequence of compact subsets of A c such that K n is contained in the interior of K n + and A c-limgK n (thus any compact subset of A c is contained in K n for n sufficiently large).Denote by u n the restriction of u on K n.Following Section 2, we can define for every n the additive functional A un A un' and these additive functionals form an increas- ing sequence.We then set A limTA.
It is easy to verify that this definition does not depend on the choice of the sequence (Kn).
Clearly AUoo may be infinite.However, on the event {50 Cq A-}, we can find a (random) integer [.J n--1 If Up(t,x)-Nt, x(f 0 NAp O), then Up is a solution of (4) in (-oc, 0)x N d (of.Subsection 2.5).By comparing Up with the function Nt, x(3,0NA-fi O) near points (O, Yn) and using the fact that the sequence (Yn) is dense in A, one easily concludes that the trace of Up is (A, 0).We claim that, for a suitable choice of the numbers rP n, infinitely many of the functions Up must be different.To see this fix a point (t0, x0).From the estimates for hitting probabilities of balls found in [1]  (Theorem 3.1), if d >_ 3 and in [11] if d-2, we can choose each number rP n small enough to make o(eO # O)   as small as desired.Hence, by a suitable choice of the numbers rP n, we can construct the sets Ap so that Up(to, Xo) tends to 0 as pc.This completes the proof.
[:l 4. The One-Dimensional Case In this section, we prove Theorem 4. We will rely on Theorem 1, although a direct argument can also be given along the lines of [14].
Let t > 0, x E R.Under Nt, x, for every r > t, the measure X r is absolutely continuous and, more precisely, Xr(dy Yr(y)dy, where the process (Yr(y),(r,y) (t, cx)R)is continuous.This follows from Theorem 1 in Sugitani [18] and the connections between super-Brownian motion and the Brownian snake.
Note that there are no nonempty polar sets in dimension 1, so that any pair (A,u), where A is a closed subset of and u a Radon measure on Ac, is a possible trace.To prove Theorem 4, it suffices to verify that, if u is a solution of (4) in (-(x), 0)x R with trace (A,) then u can be re- presented in the form u(t,x) Nt, x(3,o A -)+ Nt, x(3,o f'l A ;1-exp-]u(dy)Yo(y)). (15)   As was recalled in Subsection 2.3, we can write for t r 0, u(t,x) Nt,x(1-exp-(Xr, u(r, )>) Nt, x(1-exp-] dyYr(y)u(r,y)).
Passing to the limit r0 in this formula we see that (15) follows from the below lemma.
Proof: Part (a) is easy to prove (in fact, much easier to prove than the corresponding statement in the elliptic case [14]).We know that 3'0 is compact.Hence, on the event (3'0 A q)}, the function Y---Yo(Y) is continuous with compact support contained in \A.Because 0 is a.e.not a time of discontinuity of the mapping r---3' r (see Perkins [17] or the last section of [8]), on the same event we can find r 0 < 0 and a compact subset K of \A such that, for every r e [r 0, 0], the function Y---*Yr(Y) is supported on K.Then, if r 0 _< r < 0, f dYYr(Y)u(r, Y)) f u(dy) Yo(y) + N ( Jo o VI B C O) gl ( o VI B CIA) Nt, y(Jo[t,o]OB c O) +Nt, y(-/J[t,o]FIB c