Novel Perturbation Expansion for the Langevin Equation Carl Bender Fred Cooper Greg Kilcup L. M. Simmons, Jr. Pinaki Roy
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!
Novel Peturbation Expansion for the Langevin Equation Carl Bender Physics Department, Washington University, St, Louis, MO 63101
Fred Cooper Theoretical Division, Los Alamos National Laboratory, Los Alamos, NM 87545 L.M. Simmons, Jr. Santa Fe Institute, 1120 Canyon Road, Santa Fe, NM 87501 Pinaki Roy Electronics Uuit, India Statistical Institute, Calcutta 700035, INDIA Greg Kilcup Department of Physics, Ohio State University, Columbus, OH 43210
Abstract We discuss the randomly driven system dx/dt = -W(x)+ f(t), where f(t) is a Gaussian random function or stirring force with (f(t)f(t )) = 25(t - tf), ' and W(x) is of the form gxl+ 28• The parameter 5 is a measure of the nonlinearity of the equation. We show how to obtain the correlation functions (x(t)xW). oox(t(n)))j as a power series in 5. We obtain 3 terms in the 5 expansion and show how to use Pade approximants to analytically continue the answer in the variable 5. By using scaling relations we show how to get a uniform approximation tothe equal-time correlation functions valid for all g and
5. Key Words: Langevin equation, delta expansion, nonlinear, perturbation expansion, scaling relations
1
1. INTRODUCTION Recently a,new perturbative technique, the 8 expansion, was proposed to solve nonlinear problems in physics. 1,2,3 The technique involves replacing, in a differential equation, nonlinear terms such as x 3 by x 1+2c5 and expanding this term in powers of S: xl+ 2c5 = x
L 8n (Inn.xI2)n . 00
(1.1)
n=O
We are thus able to obtain a solution to the differential equation as a power series in 8. The . Parameter 8 is a measllrEluof the nonliIlElarityHhe theory. When 8 = 0 thet1J.eory is linear and typically can be solved in closed form. As 8 increases from zero the nonlinearity turns on smoothly. Typically the 8 series has a finite radius of convergence. Since we are interested in 8 = 1,2 etc., one needs a way of analytically continuing the series obtained to large 8. To do this we will employ Pade approximants.4 The first nontrivial Pade approximant, the [1,1] Pade, requires calculating terms up to order 82 in this expansion. We will also utilize a scaling argument to obtain the correct functional dependence of the correlation functions on the coupling constant g for all values of 8. In this paper we will be studying the one-dimensional Langevin equation
dx dt = W(x) + f(t),
W( x) = gxl+ 2c5 •
(1.2)
(For this equation to be well defined for arbitrary 8 and negative x we interpret W as follows:
W(x) =gx(x 2 )c5.) The stirring force f(t) is a random function described by Gaussian statistics, i.e., it is described by a joint probability functional
P[f] =N exp [-~ 1~ dtde f(t)S(t, tl)f(tl )] with
(1.3)
J
P[f]1Jf = 1.
Choosing white noise, we have that
(f(t)f(t' )
=
J
1J f
P[f]f(t)f(t' ) = S-l(t, t'l = 28(t - t' ),
(1.4)
where 1Jf denotes functional integration. There are two ways to determine the correlations in x(t) resulting from the statistics of the forcing term. One way is to solve directly for x(t) in terms of f(t) and then use (1.4). The other is to make a (Ad9g,A S M) = A2 < x 2 > (g,M)
(4.4)
where the dimension of the parameter M is s and the dimension of 9 is thus (from eq. (4.1)) dg = -sb+ (s -1)(2 + 28). Differentiating with respect of A and then setting A = 1 we obtain
[sM 8~ +dgg ~ -2] < x 2 > (g,M) = O.
(4.5)
This equation should not depend on the dimension of M. Differentiating with respect to s one obtains: (
8 8 2 [M 8M + (2 + 28 - b)g 8g] < x > (g, M) = 0,
(4.6)
at
which implies M = 0 at b=2+28, which is (4.3). We can write (4.2) as
(4.7) where the generic structure of f is (4.8) and
y=gMb,h=-I/(8+1).
(4.9)
The scaling condition (4.3) can be written as
{2f(y) +ybP(ynl 17
. = O. b=2+28
(4.10)
This equation is an identity if we use the exact equation for f(y) or its expansion to all orders in o. However, if we calculate only to order oN we have N
fN(Y) = Lan(y)on/nl, n=O
(4.11).
and fN(Y) is not independent of M. The scaling condition then leads to the relationship
bN(y)=daN(y)/dy=O,
(4.12)
which allows usto-choose-aparticularMso that (4~3)canbe$atisfied. Clearly bN(y) is a polynomial of degree N in In(y). Equation (4.12) has N real roots, but only the smallest root is related to the N = 1 solution. We denote this root by y* N The sequence of y*N, for N = 1,2, ... ,8, is 5.63861, 3.56103, 2.95104, 2.66687, 2.50419, 2.39919, 2.32591, 2.28717,...
(4.13)
This sequence of numbers can be fit by 1.8916 + 2.9086/N + .839096/N2.
(4.14)
Now, Y'N = gMb . Solving for M as a function of y*(N) and b, we obtain for the Nth order improved calculation: 2
(X )N
where
=:
y* ] 2/b
[
fN(y*N) ,
(4.15)
f is defined in (4.11).
We notice that if we now set b = 2 + 20 we automatically get the correct analytic behavior as a function of g. The next thing to note is that
r(3/(2 + 20)) r(1/(2 + 20)) is finite as 0 --+ 00. Thus, in trying to extrapolate to large 0 using Pade approximants one should use diagonal Pade approximants which also have the property of being finite as fj --+ 00. So the strategy is to calculate to order fjN and to solve the scaling condition for Y'N' One then forms the [N/2, N/2] Pade approximant of this answer. Only at the end does one set b = 2fj +2. To order fj we have: yi = ~exP[-('E - 3)], (4,16) where IE is Euler's constant, "{E - .57 ... Inserting this in the expansion of the answer up to order 0 one obtains as the lowest order result: (X 2)N=1 = (2fj + 2){(2g)-1 exp[-( IE _ 3)]}1/(6+1) (4.17) At 0 = 1 this yields:
18
as opposed to the exact answer [.675977]1/2. g
As expected, this approximation haS the correct analytic behavior in g and the coefficient is accurate to 10%. To order S2 we obtain instead:
Y2 = ~exP[4 -7E -
-/12 _11"2)]
(4.18)
Inserting this value of y* into the [1,1] Pade approximant of the Taylor series in S of (x 2 ) we obtain: exp[(S/(S H-){7E-4+ -/12-1I"2}][4-2y"12-1I"2 +S(30~ 6-/12-11"2)] (2g)1/(5+1) [( -/12 _11"2 - I)S +2 - -/12 _11"2]
(4.19)
Again we obtain the correct analytic behavior in g and the ratio of (4.19) to the exact answer (4.2) is a monotonic function of S, denoted by R1, having a value 1 at S = 0 and 1.37779 as S .... 00. At S = 1 the ratio is 1.0128 ... showing a great improvement over just doing the calculation up to order S. The ratio of the improved [1,1] Pade as compared to the exact answer for aIlS is shown in Figure 8. As one keeps higher and higher terms in the S expansion the ratio of the improved Pade approximants to the exact answer gets better and better. In general in order to obtain the scaling-relation improved approximation to the answer we first keep N terms in the S expansion (where N is even): N
2
< x >N= M
2
L
an(y)sn. n=O We next form the [N/2,N/2] Pade approximant for that Taylor series: "",N/2 b ( )sn M2 P[N/2,N/2](y) = M2 Lm-O n y . l:~~~ Cn(y)sn
(4.20)
(4.21)
We then evaluate this at y = y*N = gM2+28 to obtain
< x 2 > N,Y*N= (y *N /g)1/(1+8)P[N/2,N/2](y*N)'
(4.22)
So that we can compare this answer to the exact answer we form the ratio:
R
(S) =
<x2 > N,Y*N < x2 >
(4.23) N/2 We plot in Figure 9 the ratios R 2 and R 3. By the time we get to R 3 the ratio is only 6% high at S = 00. At S = 1 we obtain
R2(S = 1) = 1.000170696, R3 (S = 1) = 1.0000039263.
(4.24) (4.25)
Thus we see that the naive S expansion series, when improved by using Pade approximants which are evaluated at y*, the value given by the scaling relation, gives an approximation to the exact answer that has the exact functional dependence on the coupling constant g and gives a uniform approximation for all S when N is sufficiently large. 19
Appendix A. Summing the Series Consider 2kg r = ~ B(k ~)2k + (1- c lcr- l) 1,1 L.J . '2 2k2(k + 1) k=l If we let
s
(AI)
(A2)
then we can write
(A3) Using the integral representation for the Beta function we have
X~:l
=
00
00
k=l
k=l
L B(k, ;)xk = L x k 1-
r dtt k- 1(I_ t)1/2 = x r dt(I- xt)-l(I- t)1/2 io io 1
1
x] 1/2 ar~tan (x 1/2) [I-x]
(A4)
=2-2[ xThus we obtain
F 1(x) =2(arcsinv'x)2 + 4
1
F (1) = 3 -
1 (
~x
)1/2
arcsiny'X - 4
(A5)
i1l"2.
Using
and integratng we obtain D ( ) -3 .;;;: £2 x - - (arcsiny'X)2 - 2 (I_x)1/2 -arcsInyx x x·
F2(I) = 3 -
i1l"2.
Finally xdF3 -F ( ) --- 1 x
dx
20
(A6)
so that
F3(x) = 8 -4(arcsinv'x)2 _ 8 [ 1 ~ x ]
1/2
arcsinv'x + 4h(arcsinv'x)
where
It(Y) = faY z2 cot(z)dz
(A7)
One can either use the integral as the definition of h (y) or rewrite it in terms of the logarithmic integral functions li 2 (y) and li3 (y) •
. h(Y) = -2i Y;_+1I2(i1l" + In 2) + iyli 2(e- 2i")-li 3(e- 2i") ;
(AS)
where
We notice that 311"2
(A9)
Sll(T=D) = -4- 7 , Next consider the sums
A=
I:
R lk =_2I: 4kB (1+k,1+k)
k=1 k+1 Using B(l + k, 1 + k) =
k=1
(AID)
k(k+1)
fl dtt k (l - t)k we obtain A=2 fa1 dt~n(l-x)-l-x-l.ln(l-x)]
(All)
where x = 4t(1 - t). After two subsequent changes of variables-t = sin2 e, and then y = cos 2e we obtain
A = -2 - 4 fa1 dy lnyy2(1_ y2)-1
(A12)
Expanding the denominator, integrating term by term, and summing we obtain: 3
11"2
A = -2 - 4(1- 4"((2)) = 2"" - 6.
(A13)
Next we need the functions: 00
HO(x) = L,B(k,k)x k , k=1 00
k
00
H 1(x)
= L,B(k,k)~2' k=1
k
00
H 2(x) = L,B(k,k)k:1' k=1 . 00 k H 4(x) = L,B(k,k)~3' k=1 21
k
H 3(x) = L,B(k,k)~2' . k=1
(A14)
Using the integral representation for B(k,k) we have that (Al5)
Changing variables via 2
z2
y-y = -
. 4
we obtain
H1(x)
=~ 4foldZZ-~(L- z2}~1/~ln(l ~ x;) =[1r ~ 2arccos(~v'x)]2
(A16)
H1(4) =1r 2 . Using
x
dHl(X) _ H( ) dx x,
we obtain (Al7)
From we obtain after integrating: H 2(x) =2 + 2;2 _ 21r(4 _ x)1/2 x -l/2 + 4(4 _ x)1/2 x -l/2 arccos(~v!x)
+ 16 arctan[(4 _ x)1/2 x -l/2] arccos!'v!x) _ (~) arctan2[(4 _ x)1/2 x -l/2] x
2
x
(Al8)
_ (8;) arctan[(4 _ x )1/2 x -l/2], 1r 2
H2 (4) =2+ 2 . For H3(x) we have so
H 3(x) =
rd
io :
1 [1r - 2arccos(2v'Y)]2
(Al9)
This cannot be evaluated in terms of special functions and must be evaluated numerically. Similarly: (A20)
Appendix B. Comparison with other expansions It is instructive to compare the calculation with 3 terms in the 8 expansion to the calculation using many terms in the weak- and strong-coupling expansions. The simplest
22
place to make the comparison is for the equal-time correlation functions because we can obtain these expansions directly from the integral (1.12).
1-
) dxx 2 P(x _mx 2 gx 2(1+6) (x }eq= J~oodxj>(x) ,P(x)=exp[ 2 + (2+25)]
2
00
(Bl)
The weak-coupling expansion for this is obtained by using the Gaussian part as the· measure and expanding in powers of 9 about that. That is we write: .
__m.,2/2 ~ [ -g ]n x 2n(1+6)
-
P(x) - e
(B2)
n!· .
LJ 2 + 25 n=O ....
Performing the integrals we obtain 00
2:[-g/(2 + 26)]nr(n + 5n + 3/2)(2/m)n+6 n /nl (x2) = 2 -:0=0_ mOO
(B3)
.
2:[-g/(2 + 25)]nr(n + 5n + 1/2)(2/m)n+6nIn!
°
The weak-coupling expansion is then obtained by reexpanding this expression as a power series in 9 starting with gO. The weak-coupling expansion is of course an asymptotic series. The result of keeping 5 terms in the weak-coupling expansion is compared in Figure 10 with the exact numerical evaluation of (3.5). We notice that the weak coupling expansion breaks down at very small values of g. Taking a Pade approximant does not improve this result, as the denominators have poles at small values of 9 preventing an effective analytic continuation to large g. For the integral (3.5) one can also perform a strong-coupling expansion by treating the Gaussian part as a perturbation. That is we write:
~ (_mx 2/2)n P_( x ) = exp [_gx2+26] 2+26 LJ n!
(B4)
n=O Performing the integrals we now obtain 00
. (x 2 ) =
2:( _m 2/2)n[(2 + 25)/g]n/(1+6)r((2n + 3)/(2 + 26))/n! -0 2 + 25) 1/(1+6) ~n_~
( 9
I:
_
(B5)
(_m 2/2)n[(2 + 26)/ g]n/(1+6)r((2n + 1)/(2 + 26))/n!
n=O Reexpanding this in terms of powers of (1/ g)1/(1+6) one obtains the strong-coupling expansion. This expansion is convergent. Pade approximants to the strong-coupling expansion are a very accurate representation of the answer except for quite small g, as seen in Figure 11. Unlike the weak-coupling expansion and the 5 expansion, the strong-coupling 23
expansion exists only for zero-dimensional field theory (ordinary integrals as opposed to functional integrals). For dimensions greater than zero the strong coupling expansion must be regulated, for example by introducing a lattice because it is a singular perturbation theory in derivative terms. Thus in higher dimensions the strong-coupling expansion introduces a new parameter (the cutoff or lattice spacing) into the theory which is present even after mass and coupling constant renormalizations. 8 In particular, a strong-coupling expansion for the correlation functions (1.7) does not exist in the continuum. The lattice regulated strong coupling expansion for the path integral (1.7) is discussed in Ref. 6. On the other hand, the {j expansion, after being analytically continued using Pade approximants, works uniformly in both weak- and strong-coupling regimes and does not require the introduction of a lattice regulator in higher dimensions.
24
REFERENCES 1. C. M. Bender, K. A. Milton, M. Moshe, S. Pinsky and L. M. Simmons, Jr., Phys. Rev. Lett. 58, 2615 (1987). 2. C. M. Bender, K. A. Milton, M. Moshe, S. Pinsky and L. M. Simmons, Jr., Phys. Rev. D37, 1472 (1988). 3. C. M. Bender, F. Cooper and K. A. Milton, Phys. Rev. D 39,3684 (1989). 4. G. A. Baker, Jr. and P. Graves-Morris Pade Approximants, Part I: Theory and Pade Approximants; Part II: Extensions and Applications, Encyclopedia of Mathematics and its Applications, vol. 13 and vol. 14, Addison-Wesley Publishing Co., Reading, MA (1981). 5. R. Phythian, J. Phys. A 10, 777 (1977). 6. C. M. Bender, F. Cooper, G. Guralnik, H. A. Rose, and D. H. Sharp, J. Stat. Phys. 22,647 (1980). 7. F. Cooper and B. Freedman, Ann.Phys. (N. Y.) 146, 262 (1983). 8. C.M. Bender et.al. Phys. Rev. D23, 2976 (1981); C. M. Bender et. al. Phys. Rev. D23, 2999 (1981); G. A. Baker, L. P. Benofy, F. Cooper, and D. Preston, Nucl. Phys. B210, 273 (1982). . 9. S. Ma and G. Mazenko, Phys. Rev. Bll, 4077 (1975).
25
5. FIGURE CAPTIONS
Figure 1. Graphical representation of (2.16). Figure 2. Graphical representation of (2.27). Figure 3. Graphical representation of (2.47). Figure 4. Comparison, as a function of 8 at fixedg = 1, of the exact equal-time correlation function given by (3.2) with the [1,IJ Pade approximant (3.4), obtained from the 8 expansion through order 82. Figure 5. Comparison of (3.2) and (3.4) for fixed 8 = 1 as a function of g. Figure 6. Comparison of the exact result for < x2 > from (3.5) for m approximate result obtained as a ratio of [1,1] Pade approximants.
= 8 = 1 with the
Figure 7. Evaluation ofthe correlation function < x(r)x(O) > for 9 = 8 = 1. Langevin denotes the numerical solution including error bars. Pade denotes the [1,1] Pade of the 8 expansion Eq. (2.58) up to order 82. . Figure 8. The ratio, R b of the scaling-improved [1,1] Pade approximant to the exact answer for < x 2 > is shown as a function of 8. Figure 9. The ratios, R 2 and R 3 , of the scaling-improved [2,2] and [3,3J Pade approximants to the exact answer for < x 2 > are shown as a function of 8. Figure 10. Comparison of the first five terms in the weak-coupling expansion (B.3) to the exact evaluation of < x 2 > in (3.5), from m = 8 = 1. Figure 11. The [5,5] Pade. approximant of the strong-coupling expansion (B.5) compared to the exact evaluation of < x 2 > in (3.5), for m = 8 = 1.
26
/
• • • •
.. . . .f(5)
..
2 a.+ 1 factors f( Z i)
.------..--
.• ••
s
Figure 1
n
(a)
z
y 2~ +1
2 ex. +1
(b)
2 k+1lines
ex.. k loops
~.
Figure 2·
kloops
Y2 z1
z3 • • •
•• . ..
.• •
• • •
Y3
(a)
f(t)
(b) t
Ct-k-1 Z -loops
2 k + 1 lines
~-k
Y-loops
(c) • • •
•
. 2 k lines
Ct- k z -loops
~-k
Y-loops
Figure 3 .
g=1
0.8
G2 (t,t)
0.6
0.4
L..-L--'---'--'--'--.l-.J'--I.---'--l--'-..L-'--'--'---'--'--'--.L.......J'--I.--'--'---'
-0
2
3
Figure 4
4
5
2
1.5
0.5
4
6 9
-0.5
Figure 5
8
10
T.o
m
,
=/) = 1
,
.9
\ \
.8
, \ \
G2 (t,t)
\ \
.7
[1,1~/[1,1lo
\ \
.6
\i \
,,, J' /) expansion ,
.5 .4
--- --------
.3 .2 0
3.0
Figure 6
g
5
. .. - ... - . .. -... - ......... -... -. 1
.. - ... - ... _... - ...
- ... - .... - ........
Quadratic
. -. . . -. .
-.
-
A
~
0
~
x ~
\0'
.5
~
x
V
.2
,,
,,
Langevin
,,
\
Linear
\ \
\ \ \ \ \ \
.1
\
-0
.2
.4
Figure 7
.6
.8
1
1.35 1.3
1.2
1.1 10.5
0.0
Ii Figure 8
100
1.08
1.06
1.04
1.02
1.0 100.0
,
Figure 9
m=li= 1
1 Exact
y
0.8 5 terms weak-coupling
0.6
<x?-> 0.4
0.2
0.05
0.1
g
Figure 10
0.15
0.2
m =0= 1
1.4 [5,5] Pade of strong coupling
1.2
<X2>
1
/
0.8 0.6 0.4
Exact
0.2
1
2
3
g
Figure 11
4
5
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