A
simole model for DLA
M.Anshelevich
Pr. Nikolai Makarov
Abstract
A very simple model of a growing aggregate is considered. It is governed by a
parameter 11, and it turns out that for 11 < = 1 the system grows uniformly while for 11 > 1
the system grows almost exclusively in one direction.
Introduction
Consider the following model: we have N parallel sticks of integer length, and at
integer moments exactly one stick grows by 1, with the probability of k-th stick growing
proportional to a fixed power 1'1of its length, properly normalized. Equivalently, we
consider sequences of points of the first quadrant of Rn such that on each step exactly one
of the coordinates of the point gets incremented by 1, and the probability of k-th
coordinate being incremented is
11
(1)
r?LX~
k=1
The behavior of the system depends on the value of 11, specifically
11>1,11=1,11<1.
The (slight) similarity with the standard DLA model becomes
the sticks as growing radially away from the center a circle.
it is different
apparent
for
if we consider
We consider the following question: start with point (1,1, ... ,1). Will the coordinates
grow uniformly, or will random perturbations result in predominantly
one of the
coordinates growing? For example, for 11 = 00 the coordinate which has grown on the first
step will continue growing, while for 11=0 all the coordinates will grow at equal rate.
It will turn out that the coordinates will grow uniformly for 11 S; 1 and only one of the
coordinates will grow for 11 > 1.
Results
Define the transition
probability
Vex) on the first quadrant
of Rn as in formula (1):
x11
(1)
V k (x) =~,
k= I ,2, ..,N
LX~
k=l
Let Pt(x,y) be the probability of getting from initial point x to point y in t steps. Then P
satisfies the recursive equations
N
(2)
Pt+l (x,y) = I
PI (x +
ek,
y). V k(x)
k=1
and
N
(3)
= Ipt(x,y-ek)·
Pt+1(x,y)
Vk(y-ek)
k=!
Define the first and second momenta
(4)
M
(t, x) = I
(y - x) . P, (x, y)
y
;;
(5)
0 (t,x)
= Illy
- xii· p (x,y)
y
Then the following
(6)
difference
M(t+l,x)=
equations
hold:
I(y-x).Pt+!(x,y)=
N
= I(y-x).
IPt(x+ek,y)'
Vk(x) =
k=l
y
N
=
I II
k =1
y - (x + ek) + ek I· Vk(x)· P, (x + ek, y) =
Y
N
= I
Vk(x)· M(t,x + ek) + Vex)
k=1
N
(7)
M(t+l,x)=
I(y-x+ek)·IPJx,y)'Vk(y)=
k =1
= I[(Y
Y
- x)· P, (x,y)·
= M(t,x) + I
± Vk(y) + P, (x,y)·
k=l
V(y)· P,(x, y)
y
ek . Vk(y)] =
t\
= ~]Iy-
x112. I
p\(x +ek,y)' Yk(x) =
k =1
:\
= IIlly
k=1
- (x + ek) +ekI12.
Yk (x)· PI(x + ek,y) =
Y
N
N
= I Yk (x)· D(t, x + ek) + 2I Yk(x)· Mk (t, x + ek) + 1
k =1
k =1
N
(9)
D(t+1,x)=
IIII(y+ek)-xI12.p,(x,y)·Yk(y)=
y
k=!
II
N
=
y
Yk (y). [Ily- xl12 + 2(y- X)k+ 1]' PI (x,y) =
k=l
N
Ily- xl12. PI (x, y). I Y k(y) + 2(Y (y), (y - x)· PI (x, y)) +
II
±Y
= ~- + P \ (x, y).
k=]
k
k=!
(y)
I
= D(t,x)+2I(Y(y),(y-x)).PI(x,y)+1
y
1l = 1
N
Denote IxlT)= I x~
k=!
For 1l = 1 equation (4) becomes
1
M(t + 1,x) = M(t,x) + --I
Ixil +
IxL+ t
= (1 + _1_).
y' PI(x,y) =
t
y
Ixll'+t
M(t,x) + _l_x
Also, M(O, x) = 0
Hence M(t, x) = I:L x is the solution
Therefore equation (6) becomes
=
1
D(t+l,x)=
D(t,x)+2,
xl!
D(t+l,x)=D(t,x)+
·22(y,y-x).Pl(x,y)+1
Ixl!2+ t J.O(t,X)+
Substituting
the fomlula
Ixll2+ t (x,M(t,x»)+1
for M, get
IxL2+ t (0(t,x)+IIXll2 IxlJ tJ+l
D(t+l,x)-O(t,x)=
= a(x)t2 + b(x)t + c(x)
c(x) =
0(0, x) = 0
2a(x)t
+ (a(x)
(2a(x)lxIJ
y
2
.I[lIy-xIl2+(x,y-x)].PI(x,y)+1
Ixl! + t y
O(t+l,X)=(I+
If D(t,x)
+t
Ixll + t
+ b(x» = ~(a(x)t2
+ a(x) + b(x»)t
+ (a(x)
a(x) + b(x)
= 1 - b(x) -1{2a(X)lxt,
+ a(x)
21:::':
then
+ b(x)t
+ b(x»lxl!
Ixl! tJ+
+ IIxl12
= 2b(x)t
1
+ 211xl12t + t + Ixll
IxL
=0
IxL
1-
b(x) =(21x11a(x)
{a(x).
+ 2) = 2 + 211xl12
=
a(x)
IIxll2
+ IxL
+ 1)
IxL . (lxL
b(x) =
D(t
Ixl~ -llxl12
Ixl! .(lxL + 1)
x) =
,
Ilx//2+ IxL . t2 + Ixl~ -llxW
.t
Ixl] . (lxL + 1)
IxL . (lxL + 1)
In particular
for x lying on the main diagonal
x = (m,m,oo.,m)
t
M(t,x) = (-,-,
t
N N
D(
t
...,-)
N
)
m +1 2
m - m
m +
t
1 (
t, x = N m + 1 t + NN m + 1 t = m + -k'.JN
1- N
)2 + 1 +
1
N;n
t
t "" (
.IN )2
For m = 1
2
1--
t
D(t,x)=--j' 1+N".JN
(-
)2
t
+ __
I t;::;:2-( .IN
1+"N
N.1
)2
More precisely, if
teN)
--
N
~
N~
00
ffi )' -. 1
as
as
00
D~(
then
'
N-. ~
This means that on the average the coordinates of the point are uniform
Indeed,
x112.Pt(x,y) = II[(Yk
D(t,x) = L]Yy
y k=1
- xk _~)+~]2.
N
N
=-+II
t2
N
t2
N
Y
N [(
Yk-Xk--
k=J
t )]2
N
·Pl(X,y)~-,
Pt(x,y) =
with equality achieved only if Yk- Xk = ~
N
It follows that for 11 < I the coordinates also will be uniform
U2.-l
For
11
> 1, in order to use the equation (8)
N
D(t + 1,x) = I
k =1
N
Y k(x)D(t,
x + ek) + 2 I Y k(x)Mk (t, x + ek) + 1
k =1
We make the following approximation:
instead of equation (6) we consider the corresponding partial differential equation:
:\
M (t
+ 1, x) =
+
V (x)
I. V
k
(X) . M (t, x
+ ek ) Q
k =1
N
M(t
+ 1,x) - M(t, x) = Vex) +
(9)
aM
-(t,x)
at
(9)
aM
-(t,
at
L Vk(x)· (M(t, x + ek) k=l
aM
N
I
N
x) -
k=l
L
Vk(x)· - -(t,x)
k=l
aXk
Q
aM
(x)· (t, x) = Vex)
aXk
Q
+
= Vex)
Vk
dM
-(t(s),x(s))
= V(x(s)), where
ds
dt
-=1;
ds
dx
_k =-Vk(x(s))
ds
Hence(s)
d(M + x)
ds
= 0
xT]
Vk(x) = _k
But
IxlT]
Hence the integral curves of the equation
differ only in parametrisation
dXk
-
__
dXk
-=-xk
d't
Uk (x) - xk
T]
-X~T]dXk = d't
__
from those of
T]
- -Uk (x('r)),
d't
for x
(TJ
-:j:.
1)
= 't + C
l_xl_T]
1 -TJ k
k
I
N
with
s('r) =
k=l
Xk ('r)
Thus finally
N
with't
N ote t h'elt l' - -- 1
1-
Ck
-
L M J-r)
given by t('r) =
Ci --
'lk
(1\,1
+ xk) 'I-T]
-
C k' hence'
TJ
1 (M k + Xk)
.
I-TJ
--
and Ck = Ck (x)
k=l
d-T]
-
I (M i + Xi
1-11
--
)l-T]
M(t, x))
-7
This is true, for example,
1
1
1-yt
Let a =
T] -
-
Xi -cx
For x
= (m
-(X
1-1] -
I-yt
l-yt
1
1
(M k
--
(M k + Xk )1-'1
1-1] - Xi 1-1] _-
Xk
1
1-1] ---x
--xk
Xk
for t = 0, hence
+xk )1-1]
(M i + Xi
-
---
I-yt
(M . +x
I
)1-1]
I
)1-1]
I
= (M k + Xk
)-n
(M i + Xj
-
)-CX
+ l,m,m, ... ,m)
m-a - (m + I)-U =
+
(Mk
mru
- (Mi + m + I)-a and t
= M] + (N
1
Mk
= ( ma
1 - (m +1
It
+
+ m
1 +
(M1
It
m(m +
<--------m=
1
It -
ItmU JXx
((mmCX(m+
+
((m +
m <
JXX
-m
1)
It - mCX)XX
For m = 1
by C(ll)
Denote the above constant
Hence
> t - (N -l)C(T])
M1
Since D measures
D(t, x+ek)
non-unifonnity
of coordinates,
> D(t,x) . Thus
I
N
D(t + 1,x)
for x = (1, I ,... , I)
=
I
N
V k (x)D(t, x +
k=1
I
ek)
+2
V k (x)Mk (t,x +
k=1
N
D (t + 1, x) ~ D (t, x) + 2
V k (x)
Mk
(t,
X
+
ek
)
+1
k=1
But it has been shown
Thus D(t + 1,x)
> D(t, x) + 2t - 2 NC(ll)
Therefore
(-1
D(t,x)
Thus if
> 2 ICi - (N -1)C(T]»
i=1
>
t2
-
t - 2t(N -1)C(T])
ek)
+ 1
-I)Mkj
teN)
--
--7 00
N
D(t,x}{z
But
t2
as
--71
N --7 00,
as
N --700
is the maximum
of D. Indeed,
y
with equality achieved
only if exactly one of
(Yk -
Thus in this case exactly one of the coordinates
xk)
is non - zero
is growing
Conclusions
We see that the system does indeed exhibit qualitatively
depending on whether 11is greater or less than 1.
different
behavior for