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31 Laplace Equation and the Brownian Motion
31 Laplace Equation and the Brownian Motion 31.1 The Discrete Case Let ZNdenote the lattice of points x = (XI, . ..,x N) in RN whose coordinates xi are integers (20or
such that the image of two neighboring integers is a pair of neighboring points. If the interval in which this map is defined has a finite lower limit, the path has a starting point; if the interval has a finite upper limit, it has an arriving point. We shall now look at a “particle” moving over the lattice ZN(the reader may prefer to think of a “ piece moving on an infinite N-dimensional checkerboard). The motion of the particle is effected as follows: Before each move we throw a dice having 2N faces, on each one of which is labeled an N-tuple, precisely the set of coordinates of one of the neighbors of the origin. By translating the result from the origin to the point where the particle happens to be at present, we discover at what neighboring point it must be next. Starting with a position x, successive dice throwing (or trials) yield the path followed by the particle. Its motion, under the rules above, is called random walk. An important feature of random walk is that each trial is independent of the preceding ones. Let us denote by x(0) = x , x(l), . . ., x(n), the successive positions of the particle. We may introduce the transition probabilities p(n, x, B ) : B is a subset of ZNand p(n, x , B) is the probability that the particle starting at x(0) = x finds itself in the set B at the nth step, i.e. that x(n) E B. If B consists of a single point y, we write p(n, x, y). It is natural to define p(0, x, y ) as one if x = y and as zero otherwise. Note that ”
(31.1) p(1, x, y ) = 1/2N
if x and y are neighbors, 294
0 otherwise.
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The function B + + p ( n x, , B) defines a positive measure, with total mass one, on ZN:It is indeed a probability. If the distance between x and B exceeds n, we have p(n, x , B) = 0; in other words, y ~ p ( nx,, y ) has compact (finite) support (easy to describe). Let 52 be a subset of ZN.We shall assume R to be connected, which means that any two points in Cl can be joined by a path entirely contained in R. A point of R will be called an interior point if all its neighbors also belong to R, a boundary point if this is not so. A (say, vector-valued) function f in R is said to be harmonic if its value a t any interior point x E R is equal to its average over the set of neighbors of x : (3 1.2) Note that the right-hand side of (31.2) is nothing else but
The operator f HPf is often called the (one-step) shift or averaging operator. A real-valued function f is called superharmonic if f ( x ) 2 Pf(x) at every interior point x ; nonnegative superharmonic functions are called excessive functions in the theory of Markoff processes. It is obvious that the maximum principle holds for complex-valued harmonic functions in Q. Also, any harmonic function in the whole lattice ZN which is bounded must be a constant. We may set (31.4)
A
=
-2(I - P ) ,
where P is the shift operator (31.3) (and I the identity); A is the discrete Laplacian. By definition, a function f is harmonic if Af = 0. Suppose now that the set R is bounded, and let aR denote its boundary. Let g be a real-valued function defined in an. The Dirichlet problem in the discrete case is the problem of solving (31.5) (31.6)
Au
=0
in the interior points of R, ~
=
gonan.
This problem has a remarkable probabilistic interpretation, and so does its unique solution! Indeed, we may interpret g as a p a y o f : If our particle, starting from a point x in the interior, hits the boundary for the first time at the pointy, we have winnings or losses equal to g ( y ) [whether it is a gain or a loss depends on the sign of g ( y ) ] .The value of the solution u at x will then be the value of the expected winnings (or fosses) at x , as we are now going to show.
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BOUNDARY VALUE PROBLEMS
[Chap. I11
Let B be an arbitrary subset of the boundary dR. If x is an interior point, we define m,(B) as the probability that the random walk of the particle, starting at x , hits the boundary for the first time at a point y belonging to the set B. It is immediately seen that m, defines a positive measure on dR, with total mass 5 1. Let u s fix B and consider the nonnegative function x Hm,(B); it is defined at the interior points of R but we may extend its definition to boundary points x by setting it equal there to one if x E B , to zero otherwise. It is clear that we have, if x is an interior point,
which shows that m,(B) is harmonic. Then take B =aR: When x is a boundary point, we have m,(aR) = 1, and therefore rn,(dR) is a harmonic function in R, identically equal to one on the boundary. By the maximum principle, it must be identically equal to one throughout R. This implies that the total mass of the measure M, on 8R is one, for every x E R. Now, in order to obtain the solution u of (31.5)-(31.6) for arbitrary g, it suffices to take (3 1.8) Indeed, u is harmonic as a finite sum of harmonic functions, and it takes the value g(y) at a boundary point y , because of the properties of the harmonic measure m, [cf. (29.7)]. Because of our definition of m,({y}) (the probability that the particle first hits the boundary at the point y ) , (31.8) defines u(x) exactly as the mathematical expectation at the point x of the payoff g. We may interpret the values of any function f on ZN as coefficients of a Fourier series. Let us set, indeed, (31.9) where fj is the variable in the N-dimensional torus T N : 8 = (:
p(0, x, O) = eie‘ ,
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and on the other hand, by the total probability formula, for n 2 1, (31.12) p(n, x, y )
=
1
Ix-x'I = 1
p(1,
X,
1
x ' M n - 1, x ' , Y ) = -
2N
1
~ x - x '=~I
P(n - 1 %x'. Y L
whence 1 p"(n, x, 0) = -
(31.13)
p"(n - I , x', e).
2"x-,q=1
Taking n = 1 and applying (3 I . 1 1) yields, since I x - x ' ( x.8fBjforsomej, 1 S j S N ,
=1
implies x' * 6 =
p( I , X, e) = eie "a@),
(31.14)
'
where we have set (31.15)
If we return to (31.13) and apply induction on n obtain
=
1, 2, . .., we immediately
p(n, X, 6) = e i e .xw(e),
(3 1.16 )
and (31.11 ) shows that this is also true for n = 0. Applying now the Fourier inversion formula (31 .lo) yields
Let us then consider the sum of all the powers W,n = 0, 1, . . . , ( I - @)- ; it is clear that it has a pole at 8 = 0; the order of (1 - 0(0))-' at that pole is 2 , in the sense that (31.18) for 8 - 0 ,
[(el =(0:+...+e,Yi2] 2 3, thefunction (1 - a)-' on the torus
(1 - @ ( 0 ) ) - ' - 2 N 1 8 1 ~
We derive from this that, provided N T Nis integrable, and therefore the function (31.19)
t m
G(z) =
n=O
p ( n , z , 0 ) = ( 2 7 1 - ~ e""(1 - @(e))-' d0 T"
is bounded. If we apply the total probability formula (3 I . 12), we immediately obtain
(3120)
- +AG(x)
=6
(~),
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BOUNDARY VALUE PROBLEMS
[Chap. I11
where 6 is the analog of the Dirac measure: It is equal to one if x = 0 and to zero otherwise. Thus C is harmonic in the complement of the origin. Let us also denote by G the convolution operator (31.21)
=cG(X-Y)P(Y)
G:p-G*W
Y
(defined, say, on functions-which are the same thing as measures, on ZNwith compact support). Then (3 1.20) can be rewritten in the form (3 1.22)
( I - P)G = I,
the identity operator,
i.e., G = ( I - P)-'. If we look back at (31.15) and (31.19), we see that @(8) is is the one of the Fourier transform of the shift operator P, and (1 - @(O))-' G. The usefulness of (31.19) is that it shows readily that G(z) is a bounded function when N 2 3. One can also easily derive from it that (31.23)
when
1x1
- +oo,
G(x) - c N I x I 2 - "
[we leave the derivation as an exercise for the student: The basic idea, of course, is to exploit (31.18) and to make a change of variable 8 = IzI -'8' in the integral in (31.19)]. The properties just given show that G is the analog of the Newton potential (see Example 30.1). The presence of the factor 3 in (3 1.20) is merely due to our presently adopting a different convention (which, actually, is the one commonly adopted in potential and probability theories). We may of course define the potential in Z" of a measure (i.e., function) with compact support, p, by the formula (3 1.24)
U '
=G
*p
(cf. Definition 30.3).
It is seen at once (cf. Proposition 30.2) that U p is the unique solution of the inhomogeneous Laplace equation (31.25)
-+AU' = p
(in Z")
which tends to zero at infinity. If we rewrite (31.24) in the following fashion: (31.26) we obtain the probabilistic interpretation of the potential. Again, we must regard the given function p ( y ) as a payoff. The nth term, (31.27)
1 Y
x, Y)P(Y),
is the mathematical expectation of winnings .uat the nth trial. Equation (31.26) says therefore that U'(x) is the total mathematical expectation (or, equiva-
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lently, the mean value) of the payoff during a random walk of the particle starting at the point x. We conclude the first half of this section by computing the quantity (3 I .27) when p ( y ) = I y l’, the square of the Euclidean norm of y E ZN.The result will give us the mean value of the square of the distance (from the origin) reached by a particle in a random walk started at x. Let us denote it by d’(n, x). For n 2 1, we have (3 1.28) Of course, we have d’(0, x) = 1 x 1.’ By induction on n, we derive from (31.28) that (31.29)
d’(n, x)
=
+ n.
1x1’
Indeed, (31.28) implies
d’(n, x)
=n -
I
+ (2N)-’
2
IX’(2
[x-x’l= 1
But because of the obvious symmetry, (x-x’l=
1
I x’ 1’
= N(
I x + el l2
I
3- x
- el 1)’
= 2N(
where el is the unit vector along the x’-axis. In $3 1.2, we shall make use of (3 1.29) or rather of its consequence (which is also a direct consequence of the central limit theorem) that the mean value of the distance reached at the nth step by a particle in a random walk is of the (for large n). order of
fi
31.2 The Continuous Case Instead of dealing with the lattice ZNin $3.1, we could have dealt with any one of its homothetical images hZN (the points of hZN are obtained by multiplying those of ZN by h). Then, by going to the limit as h + +0, we should arrive at the concept of a continuous random walk. However in the process of going to the limit, we should not lose sight of the fact that the notion of distance covered during the random walk must not be allowed to dissolve! Specifically, the distance covered during a random walk in ZN must be essentially the same as the one covered in a walk “of same duration” in hZN. But we cannot, really, talk of the distance covered during a random walk, we can only talk of its mean value which, by the remark concluding Thus, if we wish to divide the length of $31.1, we know to be of the order of the steps by h-’, we must divide the “duration” of each step by h - ’ ; i.e.,
&.
300
BOUNDARY VALUE PROBLEMS
[Chap. I11
we must multiply the number of steps effected in a given time by h-’. The switch to a “temporal” terminology is not accidental: the variable n, the number of steps, in $31.1, is now going to be replaced by the time variable t , i.e., the duration of the walk. For reasons of convenience in the notation, we shall divide the duration of each step, not by h-’, but by Nh-’. Thus we return to the total probability formula (3l.12), but now in the lattice hZN and according to the preceding remarks, we rewrite it as follows: (3 1.30) [thus we have written t instead of (n - l)h2/N]. Let us denote by e j the unit vector along the xj-axis ( , j = I , . . . , N ) . We subtract p(t. x, y ) from both members in (31.30) and divide the result by h’/N. We obtain
Now we go to the limit as h + +O. We easily obtain (3 1.32) We see that the transition probability p(t, x, y ) is a solution of the heat equation, in the half-space t > 0 (with +A substituted for A). For fixed t > 0, we must have p(t, x, y ) -+ 0 as I x - y I .+ + 00, since this is true in the discrete case. Furthermore, (3 1.33) p(0, x, y ) = 6(x - y ) ,
the Dirac measure (on the diagonal).
These conditions determine p(t, x, y ) uniquely (see $6.I , also Example 44. I). We have p(t, x, y ) = p ( t , x - y ) , where (31.34)
p(t, z) = ( 2 n ~ ) - ”exp( ~ - 1 zI2/2t).
Here t > 0; when t -,+0, p(t, z ) converges (in the distributions, or in the measures sense) to the Dirac measure at the origin 6. If B is a Lebesgue measurable subset of R”, the probability that the particle, presently (i-e., at some given time which we take to be the origin) at the position x, will be, at time t > 0, at some point belonging to B is equal to (31.35)
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A probability distribution of the kind (3 1.34) is called normal or Gaussian. It is spherically symmetric, indicating the perfect isotropy of the continuous random walk (which is called Brownian motion. in honor of the botanist Robert Brown who observed the rapid oscillatory motion” of pollen grains suspended in water, in 1827). The graph of p ( t , z ) (for fixed r > 0) is the usual bell-shaped surface. with maximum at z = 0. The fact that, for arbitrary x, the probability distribution is equal to p ( t , x - y ) emphasizes the fact that the motion is translation-inuariant. Note also that, according to (31.35), past events have no influence on the future motion. One notable difference with the discrete case is that the probability that the particle finds itself at any given individual point. at some time t > 0, is always zero. The particle, presently at the point x, will at random pick up a path-according to the law (31.35). Paths are simply continuous mappings ofR, = ( t E R ; t 2 0 ) into R N ; with every “starting point” x we may associate a probability distribution. i.e., a positive measure with total mass one, on the space of all paths originating at x, but we do not go into this here. “
The probabilistic approach to the Dirichlet problem can now easily be guessed. Let R be a bounded open set in RN,dR its boundary. We introduce the time offirst exit T of a particle starting at x E R : T is the minimum of all numbers t such that x(t) E RN\R.Next we define the escape probability m,: It is a measure on the boundary dR. To every (say, Borel) subset A of 22 it assigns the number which is the probability that X ( T ) E A . If now g is an arbitrary continuous function on dR, and if we set (3 1.36) we shall have (31.37)
AM = O
(31.38)
u =g
i nR, on dR.
A few facts have to be established and the meaning of (31.38) must be made clear. First of all, we must show that the time of escape is finite, or rather (more in keeping with the standpoint adopted here) that the particle escapes from the set R with probability I . Actually, it suffices to prove that the particle escapes from an arbitrary ball, B R ( x ) ,centered at x, with probability 1. Let us show that the probability that the particle remains inside the ball forever is zero. Suppose it remains inside BR(x) until time n E Z , ; the increments x ( j ) - x ( j - I ) , j = I , . . . , n, must all have norm 5 2 R . Because of the probability law presiding over the motion, we have
P[I x ( j ) - x ( j - I ) \
s 2R] = (2n)-N’2J,
zI < 2 R
e x p ( - I ~ 1 ~ / 2 ) d z = u<~1,
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[Chap. 111
BOUNDARY VALUE PROBLEMS
and by the formula of total probability,
n n
P,[r>n]I
j= I
P[lx(j)-x(j-l)l
52R]=a;
+
which converges to zero as n + 03. (We have denoted by P [el the probability of the event e.) A very simple probabilistic argument shows that the function (31.36) is harmonic. Let B,(x) be any open ball, centered at x , whose closure is contained in R. Before hitting the boundary the particle starting at x must hit the boundary S,(x) of the ball. But then, if we want to compute the probability of hitting (at the time of first exit from R, T ) some subset of aR, we may as well forget what happened before the time of first exit from the ball B,(x), 7 ’ . In other words, we may assume that the particle starts at time 7’ from a point of S,(x). But we must specify the probability that the particle finds itself (at the time 7 ’ ) in a given subset of S,(x), A’. This must be necessarily equal to the probability that the particle hit A’ at T’ when coming from x. The latter cannot be anything but uniform on S,(x), by virtue of the symmetry of the Gaussian law. In mathematical terms, the analog of the measure m, for B,(x) is the area measure on the sphere S,(x), normalized, i.e., divided by the total area of S,(x). What we have just said means exactly that (31.39)
m,(A) = I S n - ’ [ -*
I
s,,-
I
rnx+,*(A) d i
[cf. (30.1)],
where A is a Bore1 subset of aR. In other words, the measure m, is harmonic (its measurability with respect to x is easy to verify) and this implies (31.37). Let us now look at the boundary condition (3 1.38). We know, by the considerations of Sect. 29, which led us to the notion of regular point (Definition 29.4), that things are not simple. Suppose, for instance, that R is the complement of the origin in the unit disk in the plane R2.Then it can be shown, whatever the point x in R, that is, however near x is to 0, that the probability that the particle, starting from x , will exit for the first time from R at the origin is equal to zero. The notion of regularity of a boundary point has the following probabilistic interpretation: If we denote by 7 the time o f j r s t exit from R of a particle starting at x , a point xo of dR is regular if and only if (31.40)
lim P,[t > h]
x+xo
= 0,
for every h > 0.
(in standard terminology, t converges to zero in probability as x + xo). We shall not prove this assertion (the proof is not very difficult and the student might try to find it). It is a consequence of the following assertion: xo is regular if and only if (31.41)
Vr > 0,
lim P,[x(T) E aR n Br(xo)] = 1.
x-xo
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Exercises N
31.1. Let C(z) be the potential in Z N given by (31.19). Show that, when = 2,
but that G(0) < + co if N 2 3. [Hint: In the case N = 2, observe that p(2k + 1, 0, 0) = 0 for all k = 0, 1, . . . .] 31.2. This exercise deals with the important concept introduced by Definition 31.1. A subset B of ZN is called recurrent iJ given any point x of ZN, the probability that a particle, starting at x, eventually hits B is equal to I . If x E ZN,denote by nB(x)the probability that a particle, starting at x, hits B at least once during its random walk, by &x) the probability that it hits B infinitely often. Prove the following assertion : (31.42) i f B is recurrent, nz(x) = 1 ; i f B is nonrecurrent, rc*,(x) = 0. [Hint: Express n,(x) as the sum x:zoqB(n, x , y ) , where qs(n, x, y ) is the probability that the particle, starting at x , hits the set B for the first time on the nth step, at the point y E B , and show that n f ( x ) = nB(x)nt(x).] 31.3. Prove that if N 5 2, every subset of ZNis recurrent (Definition 31. l), whereas, if N 2 3, every bounded subset ofZNis nonrecurrent. Prove that the “plane” x 3 = O in Z3is recurrent. 31.4. Prove the Riesz decomposition formula [see (30.27)]in the discrete case. That is, prove that given any excessive function f in ZN (i.e., any nonnegative superharmonic function), we have (3 1.43)
f
=
up4- f * ,
where U p = G * p is the potential of p = -$AA and f * is a harmonic function in ZN.Show thatf* is the greatest harmonic minorant off. 31.5. Let B E ZN,and nB, n: be the functions so denoted in Exercise 3 1.2. Prove that nB is excessive and that nt is the greatest harmonic minorant of IT*.Prove that p B = - +AnBis supported by B and that B is nonrecurrent if and only if 7~~ is a potential. 31.6. Suppose that B is nonrecurrent (see Definition 31.1). Prove that p B = -$AnB is the greatest charge carried by B whose potential is I1. [Then the total mass %(B) = p B ( y )is called the (discrete) capacity of B.1 31.7. Prove that, if N 1 3 , the capacity of a set B consisting of k points (k < + co) is equal to k/G(O), G(z) being the potential (31.19), and that the capacity of any infinite nonrecurrent set is infinite.
xyeB
304
BOUNDARY VALUE PROBLEMS
[Chap. 111
31.8. Let R be a bounded open subset of RN. We shall denote by a the first strictly positive time of exit from 51 of a particle starting at x. (Note that if XECR, the first time of exit z is obviously zero, but if x belongs to the boundary of R, the particle might reenter R at once and then reexit from R at a later time.) Let h be an arbitrary number greater than zero. Show that, given any point x, (31.44)
P,[z > h] 5 P,[a > h].
Denote then by A(to) the event that, in the time interval t o 5 t 5 h ( t o > 0), the trajectory of the particle lies entirely in R, and denote by P,[A(t,)] the probability of A(?,) for a particle starting at x. Prove that P,[A(t,)], for fixed t o , 0 < to < h, is a continuous function of x in RN. ( H i n t : Prove that (31.45) Show that (3 1.46)
PX[o> h] = lim P,[A(t,)], bL+O
and conclude from this that P,[a > h] is upper semicontinuous with respect to x , and that, by using the necessary and sufficient condition (3 1.40) for the regularity of xo E dR, together with (3 1.44), xo is regular if (3 I .47)
Px,[a > 01 = 0.
31.9. Use the same notation as in Exercise 31.8. Let h, t , be two numbers such that 0 < h < t, and let now A ( ? , ) denote the event that, in the time interval h 2 t 5 t , , the trajectory of the particle lies entirely inside R. Prove that, if x is any point in R”, (3 1.48)
Px[a> 0, 4 f l ) I = P,[a > OlP,[A(tl)l,
i.e., for h > 0, the events a > 0 and A(?,) are independent (observe that the occurrence of a > 0 is decided in an infinitely small time interval 0 < t < E ) . Derive from (31.48) that (31.49)
P,[O
> 01 = (P,[a > 0])2,
whence the zero-one law: P,[a > 01 is either equal to zero or to one. 31.10. Let R be a bounded open subset of RN, xo a point of its boundary. Suppose that there is a truncated spherical cone r, contained in RN\R, with vertex at xo . By combining the sufficient condition of regularity (31.47) with the zero-one law (31.49), show that if xo were irregular, there would be to > 0 such that the probability that the trajectory lie entirely inside R during the time interval 0 < t < t o would be 1. Thus for 0 < t < t o , the trajectory
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would certainly lie outside the cone r. Use then the rotation invariance of the Gaussian law to conclude that, during the open interval 0 < t < t o , the trajectory would lie outside an open ball, with strictly positive radius, centered at x o , and derive from this that xo must be regular, contrary to the assumption. 31.11. Derive from the Gaussian law the mean value of the distance covered, during a time interval 0 5 t 5 T, in a continuous random walk.
such that the image of two neighboring integers is a pair of neighboring points. If the interval in which this map is defined has a finite lower limit, the path has a starting point; if the interval has a finite upper limit, it has an arriving point. We shall now look at a “particle” moving over the lattice ZN(the reader may prefer to think of a “ piece moving on an infinite N-dimensional checkerboard). The motion of the particle is effected as follows: Before each move we throw a dice having 2N faces, on each one of which is labeled an N-tuple, precisely the set of coordinates of one of the neighbors of the origin. By translating the result from the origin to the point where the particle happens to be at present, we discover at what neighboring point it must be next. Starting with a position x, successive dice throwing (or trials) yield the path followed by the particle. Its motion, under the rules above, is called random walk. An important feature of random walk is that each trial is independent of the preceding ones. Let us denote by x(0) = x , x(l), . . ., x(n), the successive positions of the particle. We may introduce the transition probabilities p(n, x, B ) : B is a subset of ZNand p(n, x , B) is the probability that the particle starting at x(0) = x finds itself in the set B at the nth step, i.e. that x(n) E B. If B consists of a single point y, we write p(n, x, y). It is natural to define p(0, x, y ) as one if x = y and as zero otherwise. Note that ”
(31.1) p(1, x, y ) = 1/2N
if x and y are neighbors, 294
0 otherwise.
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The function B + + p ( n x, , B) defines a positive measure, with total mass one, on ZN:It is indeed a probability. If the distance between x and B exceeds n, we have p(n, x , B) = 0; in other words, y ~ p ( nx,, y ) has compact (finite) support (easy to describe). Let 52 be a subset of ZN.We shall assume R to be connected, which means that any two points in Cl can be joined by a path entirely contained in R. A point of R will be called an interior point if all its neighbors also belong to R, a boundary point if this is not so. A (say, vector-valued) function f in R is said to be harmonic if its value a t any interior point x E R is equal to its average over the set of neighbors of x : (3 1.2) Note that the right-hand side of (31.2) is nothing else but
The operator f HPf is often called the (one-step) shift or averaging operator. A real-valued function f is called superharmonic if f ( x ) 2 Pf(x) at every interior point x ; nonnegative superharmonic functions are called excessive functions in the theory of Markoff processes. It is obvious that the maximum principle holds for complex-valued harmonic functions in Q. Also, any harmonic function in the whole lattice ZN which is bounded must be a constant. We may set (31.4)
A
=
-2(I - P ) ,
where P is the shift operator (31.3) (and I the identity); A is the discrete Laplacian. By definition, a function f is harmonic if Af = 0. Suppose now that the set R is bounded, and let aR denote its boundary. Let g be a real-valued function defined in an. The Dirichlet problem in the discrete case is the problem of solving (31.5) (31.6)
Au
=0
in the interior points of R, ~
=
gonan.
This problem has a remarkable probabilistic interpretation, and so does its unique solution! Indeed, we may interpret g as a p a y o f : If our particle, starting from a point x in the interior, hits the boundary for the first time at the pointy, we have winnings or losses equal to g ( y ) [whether it is a gain or a loss depends on the sign of g ( y ) ] .The value of the solution u at x will then be the value of the expected winnings (or fosses) at x , as we are now going to show.
296
BOUNDARY VALUE PROBLEMS
[Chap. I11
Let B be an arbitrary subset of the boundary dR. If x is an interior point, we define m,(B) as the probability that the random walk of the particle, starting at x , hits the boundary for the first time at a point y belonging to the set B. It is immediately seen that m, defines a positive measure on dR, with total mass 5 1. Let u s fix B and consider the nonnegative function x Hm,(B); it is defined at the interior points of R but we may extend its definition to boundary points x by setting it equal there to one if x E B , to zero otherwise. It is clear that we have, if x is an interior point,
which shows that m,(B) is harmonic. Then take B =aR: When x is a boundary point, we have m,(aR) = 1, and therefore rn,(dR) is a harmonic function in R, identically equal to one on the boundary. By the maximum principle, it must be identically equal to one throughout R. This implies that the total mass of the measure M, on 8R is one, for every x E R. Now, in order to obtain the solution u of (31.5)-(31.6) for arbitrary g, it suffices to take (3 1.8) Indeed, u is harmonic as a finite sum of harmonic functions, and it takes the value g(y) at a boundary point y , because of the properties of the harmonic measure m, [cf. (29.7)]. Because of our definition of m,({y}) (the probability that the particle first hits the boundary at the point y ) , (31.8) defines u(x) exactly as the mathematical expectation at the point x of the payoff g. We may interpret the values of any function f on ZN as coefficients of a Fourier series. Let us set, indeed, (31.9) where fj is the variable in the N-dimensional torus T N : 8 = (:
p(0, x, O) = eie‘ ,
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297
and on the other hand, by the total probability formula, for n 2 1, (31.12) p(n, x, y )
=
1
Ix-x'I = 1
p(1,
X,
1
x ' M n - 1, x ' , Y ) = -
2N
1
~ x - x '=~I
P(n - 1 %x'. Y L
whence 1 p"(n, x, 0) = -
(31.13)
p"(n - I , x', e).
2"x-,q=1
Taking n = 1 and applying (3 I . 1 1) yields, since I x - x ' ( x.8fBjforsomej, 1 S j S N ,
=1
implies x' * 6 =
p( I , X, e) = eie "a@),
(31.14)
'
where we have set (31.15)
If we return to (31.13) and apply induction on n obtain
=
1, 2, . .., we immediately
p(n, X, 6) = e i e .xw(e),
(3 1.16 )
and (31.11 ) shows that this is also true for n = 0. Applying now the Fourier inversion formula (31 .lo) yields
Let us then consider the sum of all the powers W,n = 0, 1, . . . , ( I - @)- ; it is clear that it has a pole at 8 = 0; the order of (1 - 0(0))-' at that pole is 2 , in the sense that (31.18) for 8 - 0 ,
[(el =(0:+...+e,Yi2] 2 3, thefunction (1 - a)-' on the torus
(1 - @ ( 0 ) ) - ' - 2 N 1 8 1 ~
We derive from this that, provided N T Nis integrable, and therefore the function (31.19)
t m
G(z) =
n=O
p ( n , z , 0 ) = ( 2 7 1 - ~ e""(1 - @(e))-' d0 T"
is bounded. If we apply the total probability formula (3 I . 12), we immediately obtain
(3120)
- +AG(x)
=6
(~),
298
BOUNDARY VALUE PROBLEMS
[Chap. I11
where 6 is the analog of the Dirac measure: It is equal to one if x = 0 and to zero otherwise. Thus C is harmonic in the complement of the origin. Let us also denote by G the convolution operator (31.21)
=cG(X-Y)P(Y)
G:p-G*W
Y
(defined, say, on functions-which are the same thing as measures, on ZNwith compact support). Then (3 1.20) can be rewritten in the form (3 1.22)
( I - P)G = I,
the identity operator,
i.e., G = ( I - P)-'. If we look back at (31.15) and (31.19), we see that @(8) is is the one of the Fourier transform of the shift operator P, and (1 - @(O))-' G. The usefulness of (31.19) is that it shows readily that G(z) is a bounded function when N 2 3. One can also easily derive from it that (31.23)
when
1x1
- +oo,
G(x) - c N I x I 2 - "
[we leave the derivation as an exercise for the student: The basic idea, of course, is to exploit (31.18) and to make a change of variable 8 = IzI -'8' in the integral in (31.19)]. The properties just given show that G is the analog of the Newton potential (see Example 30.1). The presence of the factor 3 in (3 1.20) is merely due to our presently adopting a different convention (which, actually, is the one commonly adopted in potential and probability theories). We may of course define the potential in Z" of a measure (i.e., function) with compact support, p, by the formula (3 1.24)
U '
=G
*p
(cf. Definition 30.3).
It is seen at once (cf. Proposition 30.2) that U p is the unique solution of the inhomogeneous Laplace equation (31.25)
-+AU' = p
(in Z")
which tends to zero at infinity. If we rewrite (31.24) in the following fashion: (31.26) we obtain the probabilistic interpretation of the potential. Again, we must regard the given function p ( y ) as a payoff. The nth term, (31.27)
1 Y
x, Y)P(Y),
is the mathematical expectation of winnings .uat the nth trial. Equation (31.26) says therefore that U'(x) is the total mathematical expectation (or, equiva-
Sect. 311
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299
lently, the mean value) of the payoff during a random walk of the particle starting at the point x. We conclude the first half of this section by computing the quantity (3 I .27) when p ( y ) = I y l’, the square of the Euclidean norm of y E ZN.The result will give us the mean value of the square of the distance (from the origin) reached by a particle in a random walk started at x. Let us denote it by d’(n, x). For n 2 1, we have (3 1.28) Of course, we have d’(0, x) = 1 x 1.’ By induction on n, we derive from (31.28) that (31.29)
d’(n, x)
=
+ n.
1x1’
Indeed, (31.28) implies
d’(n, x)
=n -
I
+ (2N)-’
2
IX’(2
[x-x’l= 1
But because of the obvious symmetry, (x-x’l=
1
I x’ 1’
= N(
I x + el l2
I
3- x
- el 1)’
= 2N(
where el is the unit vector along the x’-axis. In $3 1.2, we shall make use of (3 1.29) or rather of its consequence (which is also a direct consequence of the central limit theorem) that the mean value of the distance reached at the nth step by a particle in a random walk is of the (for large n). order of
fi
31.2 The Continuous Case Instead of dealing with the lattice ZNin $3.1, we could have dealt with any one of its homothetical images hZN (the points of hZN are obtained by multiplying those of ZN by h). Then, by going to the limit as h + +0, we should arrive at the concept of a continuous random walk. However in the process of going to the limit, we should not lose sight of the fact that the notion of distance covered during the random walk must not be allowed to dissolve! Specifically, the distance covered during a random walk in ZN must be essentially the same as the one covered in a walk “of same duration” in hZN. But we cannot, really, talk of the distance covered during a random walk, we can only talk of its mean value which, by the remark concluding Thus, if we wish to divide the length of $31.1, we know to be of the order of the steps by h-’, we must divide the “duration” of each step by h - ’ ; i.e.,
&.
300
BOUNDARY VALUE PROBLEMS
[Chap. I11
we must multiply the number of steps effected in a given time by h-’. The switch to a “temporal” terminology is not accidental: the variable n, the number of steps, in $31.1, is now going to be replaced by the time variable t , i.e., the duration of the walk. For reasons of convenience in the notation, we shall divide the duration of each step, not by h-’, but by Nh-’. Thus we return to the total probability formula (3l.12), but now in the lattice hZN and according to the preceding remarks, we rewrite it as follows: (3 1.30) [thus we have written t instead of (n - l)h2/N]. Let us denote by e j the unit vector along the xj-axis ( , j = I , . . . , N ) . We subtract p(t. x, y ) from both members in (31.30) and divide the result by h’/N. We obtain
Now we go to the limit as h + +O. We easily obtain (3 1.32) We see that the transition probability p(t, x, y ) is a solution of the heat equation, in the half-space t > 0 (with +A substituted for A). For fixed t > 0, we must have p(t, x, y ) -+ 0 as I x - y I .+ + 00, since this is true in the discrete case. Furthermore, (3 1.33) p(0, x, y ) = 6(x - y ) ,
the Dirac measure (on the diagonal).
These conditions determine p(t, x, y ) uniquely (see $6.I , also Example 44. I). We have p(t, x, y ) = p ( t , x - y ) , where (31.34)
p(t, z) = ( 2 n ~ ) - ”exp( ~ - 1 zI2/2t).
Here t > 0; when t -,+0, p(t, z ) converges (in the distributions, or in the measures sense) to the Dirac measure at the origin 6. If B is a Lebesgue measurable subset of R”, the probability that the particle, presently (i-e., at some given time which we take to be the origin) at the position x, will be, at time t > 0, at some point belonging to B is equal to (31.35)
Sect. 311
LAPLACE EQUATION A N D THE BROWNIAN MOTION
30 1
A probability distribution of the kind (3 1.34) is called normal or Gaussian. It is spherically symmetric, indicating the perfect isotropy of the continuous random walk (which is called Brownian motion. in honor of the botanist Robert Brown who observed the rapid oscillatory motion” of pollen grains suspended in water, in 1827). The graph of p ( t , z ) (for fixed r > 0) is the usual bell-shaped surface. with maximum at z = 0. The fact that, for arbitrary x, the probability distribution is equal to p ( t , x - y ) emphasizes the fact that the motion is translation-inuariant. Note also that, according to (31.35), past events have no influence on the future motion. One notable difference with the discrete case is that the probability that the particle finds itself at any given individual point. at some time t > 0, is always zero. The particle, presently at the point x, will at random pick up a path-according to the law (31.35). Paths are simply continuous mappings ofR, = ( t E R ; t 2 0 ) into R N ; with every “starting point” x we may associate a probability distribution. i.e., a positive measure with total mass one, on the space of all paths originating at x, but we do not go into this here. “
The probabilistic approach to the Dirichlet problem can now easily be guessed. Let R be a bounded open set in RN,dR its boundary. We introduce the time offirst exit T of a particle starting at x E R : T is the minimum of all numbers t such that x(t) E RN\R.Next we define the escape probability m,: It is a measure on the boundary dR. To every (say, Borel) subset A of 22 it assigns the number which is the probability that X ( T ) E A . If now g is an arbitrary continuous function on dR, and if we set (3 1.36) we shall have (31.37)
AM = O
(31.38)
u =g
i nR, on dR.
A few facts have to be established and the meaning of (31.38) must be made clear. First of all, we must show that the time of escape is finite, or rather (more in keeping with the standpoint adopted here) that the particle escapes from the set R with probability I . Actually, it suffices to prove that the particle escapes from an arbitrary ball, B R ( x ) ,centered at x, with probability 1. Let us show that the probability that the particle remains inside the ball forever is zero. Suppose it remains inside BR(x) until time n E Z , ; the increments x ( j ) - x ( j - I ) , j = I , . . . , n, must all have norm 5 2 R . Because of the probability law presiding over the motion, we have
P[I x ( j ) - x ( j - I ) \
s 2R] = (2n)-N’2J,
zI < 2 R
e x p ( - I ~ 1 ~ / 2 ) d z = u<~1,
302
[Chap. 111
BOUNDARY VALUE PROBLEMS
and by the formula of total probability,
n n
P,[r>n]I
j= I
P[lx(j)-x(j-l)l
52R]=a;
+
which converges to zero as n + 03. (We have denoted by P [el the probability of the event e.) A very simple probabilistic argument shows that the function (31.36) is harmonic. Let B,(x) be any open ball, centered at x , whose closure is contained in R. Before hitting the boundary the particle starting at x must hit the boundary S,(x) of the ball. But then, if we want to compute the probability of hitting (at the time of first exit from R, T ) some subset of aR, we may as well forget what happened before the time of first exit from the ball B,(x), 7 ’ . In other words, we may assume that the particle starts at time 7’ from a point of S,(x). But we must specify the probability that the particle finds itself (at the time 7 ’ ) in a given subset of S,(x), A’. This must be necessarily equal to the probability that the particle hit A’ at T’ when coming from x. The latter cannot be anything but uniform on S,(x), by virtue of the symmetry of the Gaussian law. In mathematical terms, the analog of the measure m, for B,(x) is the area measure on the sphere S,(x), normalized, i.e., divided by the total area of S,(x). What we have just said means exactly that (31.39)
m,(A) = I S n - ’ [ -*
I
s,,-
I
rnx+,*(A) d i
[cf. (30.1)],
where A is a Bore1 subset of aR. In other words, the measure m, is harmonic (its measurability with respect to x is easy to verify) and this implies (31.37). Let us now look at the boundary condition (3 1.38). We know, by the considerations of Sect. 29, which led us to the notion of regular point (Definition 29.4), that things are not simple. Suppose, for instance, that R is the complement of the origin in the unit disk in the plane R2.Then it can be shown, whatever the point x in R, that is, however near x is to 0, that the probability that the particle, starting from x , will exit for the first time from R at the origin is equal to zero. The notion of regularity of a boundary point has the following probabilistic interpretation: If we denote by 7 the time o f j r s t exit from R of a particle starting at x , a point xo of dR is regular if and only if (31.40)
lim P,[t > h]
x+xo
= 0,
for every h > 0.
(in standard terminology, t converges to zero in probability as x + xo). We shall not prove this assertion (the proof is not very difficult and the student might try to find it). It is a consequence of the following assertion: xo is regular if and only if (31.41)
Vr > 0,
lim P,[x(T) E aR n Br(xo)] = 1.
x-xo
Sect. 311
LAPLACE EQUATION AND THE BROWNIAN MOTION
303
Exercises N
31.1. Let C(z) be the potential in Z N given by (31.19). Show that, when = 2,
but that G(0) < + co if N 2 3. [Hint: In the case N = 2, observe that p(2k + 1, 0, 0) = 0 for all k = 0, 1, . . . .] 31.2. This exercise deals with the important concept introduced by Definition 31.1. A subset B of ZN is called recurrent iJ given any point x of ZN, the probability that a particle, starting at x, eventually hits B is equal to I . If x E ZN,denote by nB(x)the probability that a particle, starting at x, hits B at least once during its random walk, by &x) the probability that it hits B infinitely often. Prove the following assertion : (31.42) i f B is recurrent, nz(x) = 1 ; i f B is nonrecurrent, rc*,(x) = 0. [Hint: Express n,(x) as the sum x:zoqB(n, x , y ) , where qs(n, x, y ) is the probability that the particle, starting at x , hits the set B for the first time on the nth step, at the point y E B , and show that n f ( x ) = nB(x)nt(x).] 31.3. Prove that if N 5 2, every subset of ZNis recurrent (Definition 31. l), whereas, if N 2 3, every bounded subset ofZNis nonrecurrent. Prove that the “plane” x 3 = O in Z3is recurrent. 31.4. Prove the Riesz decomposition formula [see (30.27)]in the discrete case. That is, prove that given any excessive function f in ZN (i.e., any nonnegative superharmonic function), we have (3 1.43)
f
=
up4- f * ,
where U p = G * p is the potential of p = -$AA and f * is a harmonic function in ZN.Show thatf* is the greatest harmonic minorant off. 31.5. Let B E ZN,and nB, n: be the functions so denoted in Exercise 3 1.2. Prove that nB is excessive and that nt is the greatest harmonic minorant of IT*.Prove that p B = - +AnBis supported by B and that B is nonrecurrent if and only if 7~~ is a potential. 31.6. Suppose that B is nonrecurrent (see Definition 31.1). Prove that p B = -$AnB is the greatest charge carried by B whose potential is I1. [Then the total mass %(B) = p B ( y )is called the (discrete) capacity of B.1 31.7. Prove that, if N 1 3 , the capacity of a set B consisting of k points (k < + co) is equal to k/G(O), G(z) being the potential (31.19), and that the capacity of any infinite nonrecurrent set is infinite.
xyeB
304
BOUNDARY VALUE PROBLEMS
[Chap. 111
31.8. Let R be a bounded open subset of RN. We shall denote by a the first strictly positive time of exit from 51 of a particle starting at x. (Note that if XECR, the first time of exit z is obviously zero, but if x belongs to the boundary of R, the particle might reenter R at once and then reexit from R at a later time.) Let h be an arbitrary number greater than zero. Show that, given any point x, (31.44)
P,[z > h] 5 P,[a > h].
Denote then by A(to) the event that, in the time interval t o 5 t 5 h ( t o > 0), the trajectory of the particle lies entirely in R, and denote by P,[A(t,)] the probability of A(?,) for a particle starting at x. Prove that P,[A(t,)], for fixed t o , 0 < to < h, is a continuous function of x in RN. ( H i n t : Prove that (31.45) Show that (3 1.46)
PX[o> h] = lim P,[A(t,)], bL+O
and conclude from this that P,[a > h] is upper semicontinuous with respect to x , and that, by using the necessary and sufficient condition (3 1.40) for the regularity of xo E dR, together with (3 1.44), xo is regular if (3 I .47)
Px,[a > 01 = 0.
31.9. Use the same notation as in Exercise 31.8. Let h, t , be two numbers such that 0 < h < t, and let now A ( ? , ) denote the event that, in the time interval h 2 t 5 t , , the trajectory of the particle lies entirely inside R. Prove that, if x is any point in R”, (3 1.48)
Px[a> 0, 4 f l ) I = P,[a > OlP,[A(tl)l,
i.e., for h > 0, the events a > 0 and A(?,) are independent (observe that the occurrence of a > 0 is decided in an infinitely small time interval 0 < t < E ) . Derive from (31.48) that (31.49)
P,[O
> 01 = (P,[a > 0])2,
whence the zero-one law: P,[a > 01 is either equal to zero or to one. 31.10. Let R be a bounded open subset of RN, xo a point of its boundary. Suppose that there is a truncated spherical cone r, contained in RN\R, with vertex at xo . By combining the sufficient condition of regularity (31.47) with the zero-one law (31.49), show that if xo were irregular, there would be to > 0 such that the probability that the trajectory lie entirely inside R during the time interval 0 < t < t o would be 1. Thus for 0 < t < t o , the trajectory
Sect. 311
LAPLACE EQUATION A N D THE BROWNIAN MOTION
305
would certainly lie outside the cone r. Use then the rotation invariance of the Gaussian law to conclude that, during the open interval 0 < t < t o , the trajectory would lie outside an open ball, with strictly positive radius, centered at x o , and derive from this that xo must be regular, contrary to the assumption. 31.11. Derive from the Gaussian law the mean value of the distance covered, during a time interval 0 5 t 5 T, in a continuous random walk.