A method for solving Boltzmann's equation in semiconductors
by expansion in Legendre polynomials
Citation for published version (APA):
Roer, van de, T. G., & van Someren Greve, S. C. (1983). A method for solving Boltzmann's equation in
semiconductors by expansion in Legendre polynomials. (EUT report. E, Fac. of Electrical Engineering; Vol. 83-E-134). Technische Hogeschool Eindhoven.
Document status and date: Published: 01/01/1983
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Electrical Engineering
A method for solving Boltzmann's equation in semiconductors by expansion in Legendre polynomials By
Th.G. van de Roer and
S.C. van Someren Greve
EUT Report 83-E-134 ISBN 90-6144-134-X ISSN 0167-9708 January 1983
EINDHOVEN UNIVERSITY OF TECHNOLOGY
Department of Electrical Engineering Eindhoven The Netherlands
A METHOD FOR SOLVING BOLTZMANN'S EQUATION IN SEMICONDUCTORS BY EXPANSION IN LEGENDRE
POLYNOMIALS
By
Th.G. van de Roer and
s.c.
van SomerenGreve
EUT Report 83-E-134 ISBN 90-6144-134-X ISSN 0167-9708
Eindhoven January 1983
Roer,/ T.G. van de
A method for solving Boltzmann's equation in semiconductors by expansion in Legendre polynomials /
by T.G. van de Roer and S.C. van Someren Greve.
Eindhoven: University of Technology. Fig.
-(Eindhoven university of technology research reports, IS5N 0167-9708; 83-E-134)
Met lit. opg., reg. ISBN 90-6144-134-X
5ISO 664.3 UDC 621.382 UGI650
Summary
A method to calculate the transport of carriers in a semiconductor is described. It relies on an iterative solution of Boltzmann's equation, in which an assumed distribution function (DF) in k-space
is substituted into the r.h.s. (the part that describes scattering)
after which the resulting differential equation is solved, giving a new approximation for the DF, etc. In the method described here
the angular dependence of the DF is, assuming symmetry around the
direction of the electric field, expanded into a series of
Legendre polynomials. This expansion is substituted into Boltzmann's
equation, leading to an infinite set of equations which may be truncated after a certain number, depending on the desired accuracy.
It has been found that, introducing self-scattering (i.e. adding fictitious scatterings which do not alter the DF, so that the-total scattering rate becomes independent of k) the solution of the set of differential equations can be found in the form of
integrals which are easy to evaluate numerically. The present report is devoted to the derivation of this solution.
Roer, Th.G. van de and S.C. van Someren Greve
A IffiTHOD FOR SOLVING BOLTZMANN'S EQUATION IN SEMICONDUCTORS BY EXPANSION IN LEGENDRE POLYNOMIALS.
Department of Electrical Engineering, Eindhoven University of Technology, 1983.
EUT Report 83-E-134
Address of the authors:
Group of Networks and Fields,
Department of Electrical Engineering, Eindhoven University of Technology,
P.O. Box 513, 5600 MB EINDHOVEN, The Netherlands
CONTENTS
I. Introduction
II. Legendre-polynomial expansion of Boltzmann's
equation
III.
IV. V.
The homogenous solution
Solution of the inhomogenous equation Boundary conditions Appendix A References 3 7 12 16 19 20
I. Introduction
In recent years considerable interest has developed in the calculation
of carrier distribution functions in semiconductors, owing to the fact
that with the small dimensions, high electric fields and short timescales
encountered in modern electronic devices the transport equation approach
with simple mobility and diffusion coefficients is no longer adequate. Of the methods developed to calculate the distribution function (D.F.) the most popular are Monte Carlo methods, in which the path of one or more
carriers in momentum space and in real space is simulated directly, and iterative solutions of Boltzmann's equation in which first the r.h.s.
(the collision part) is calculated assuming a D.F. and then solving the l.h.s. (the field part) to obtain a better approximation for the D.F., a
process which is repeated until convergence occurs.
Monte Carlo calculations are very time-consuming so that for device cal-culations an iterative method is preferred. The best known of these is the method of Rees [ I ] in which the time- and space-independent D.F. is cal-culated by converting Boltzmann's equation to an integral equation which is iteratively solved. The time-space-dependent D.F. is then calculated using a set of static D.F.'s at different field strengths as a basis. It is important to note that time-dependence can be included in an elegant way by introducing self-scattering, a fictitious scattering process that does not change the momentum and energy of the carriers. When the
self-scattering rate is defined such that the total self-scattering rate remains constant, each step in the iteration procedure is equivalent to a time
step in reality.
Another method is to expand the distribution function in orthogonal Legendre polynomials (this presupposes there is a symmetry axis) which leads to an
infinite set of integro-differential equations which, after truncation, can be solved iteratively. The difficulty with this method is that the equations have homogeneous solutions which explode either at zero momentum or at
infinity. Hammar [2J has found a clever method to subtract the unwanted solutions during numerical integrations of the equations. The method however becomes very complicated when a large number of equations is taken into account. With a small number of equations the details of the D.F. are lost but Hammar has demonstrated that averages like velocity and energy are still obtained with good accuracy.
In the following it will be shown that Hammar's method can be improved by the introduction of self scattering. An analytical solution of the set of equations is then possible and the problems with the homogeneous
solutions can be avoided. Also, as in Rees' method, each iteration step
now is equivalent to a physical time step. Furthermore the method is
not more difficult to implement numerically for a large number of equations than for a small number. The computational speed is similar to that of Hammar's method.
The method is independent of the scattering process considered so these will not be entered upon in the following. It is also equally well
applicable to many-valley semiconductors as to single-valley semiconductors. In a previous paper [5J results on GaAs and GaInAs have been presented. The present paper concentrates on the mathematical aspects of the method.
II. Legendre-polynomial expansion of Boltzmann's equation
Boltzmann's equation in its space-independent form reads:
af(~,t) e~ af(~,t)
at +-h-' -:a;";k-- (II-I)
Here E is the applied electric field, S(~,~') is the probability of scattering in unit time from wavevector ~ to
!'.
For each relevantscattering process there exists a separate function S. The properties of these scattering probabilities are not the subject of this paper, but it will be assumed that they are all well-behaved functions of k and k'. We then can write (I) as
af af
at
+ ~. ak=
g(~,t) - v(~)f(~,t) (II-2)where the reduced electric field ~ has been introduced. Usually the scattering processes are isotropic and v is a function of the modulus of k alone:
v (~) v (k) •
The time-independent form of (2) can be solved iteratively by assuming a distribution function f(~), calculating g(~) and then solving the resulting linear differential equation. Following Rees [IJ, we can introduce a self-scattering probability, describing fictitious self-scatterings from k to k. This probability is chosen such that the total scattering out of k becomes constant. The (time-independent) equation then reads:
df(~)
F · - - - +
- dk (II-3)
As Rees has shown each step in the iteration now is equivalent to a
physical time step Ifr provided r is large enough. This is easy to show as follows: in the nth iteration step we have
df(n)
F. - - - + dk
However, when the time dependence is included we must have:
Subtracting the last two equations we find
F
+ -r
1
Now when we put r
=
6t
then in the limit fit + 0 the second term on the right becomes vanishing small and the expected result follows.To solve eq. (3), with g (k) a known function, different courses can be s
-followed. E.g. Rees [IJ converted (3) to an integral:
which was evaluated numerically. We will follow the approach of expanding
f(~) in a series of orthogonal Legendre polynomials which was applied by Hammar [2J to eq. (2), i.e. without involving self-scattering. The infinite
set of equations which obtains, was, after truncation, solved by Hammar
numerically. The problem of numerical instability that results of the homogeneous solutions which are unbounded at k
=
0 or k= "',
were solved by him in a very ingenious but complicated way. In this paper it will be shown that, applying the Legendre-polynomial expansion to eq. (3) an analyticalsolution is possible which offers the same computational speed as Hammar's
method and which has the added advantage of describing the time evolution of the distribution function following a field step.
Let us now introduce a polar coordinate system (k,8,~) in k-space, with the
polar axis in the direction of the electric field. Then, all scattering processes being isotropic, we may assume that f possesses circular symmetry
so that we can write f(~) as
f(~)
=
f(k,cos8).Making the coordinate transformations
r
F
r k, s = cosSwe can write (3) as: df(r,s) s dr 2 1-s + -r df(r,s) dS + fer,s) (II-4) (II-5)
The next step is to apply a Legendre-polynomial expansion to f and g : s f(r,s) =
L
f (r)P (s)n n g (r,s) s =
L
g (r)P (s). n no o
The first few Legendre polynomials are given in Table I.
Table 1. n P (s)
o
2 3 4 5 6 n 5 H3s2-1) 1<553-3s) 1 4 2 "8(35s -30s +3) I 5 3 '8(63s -70s +15s) I 6 4 2 16(23Is -315s +1058 -5)They satisfy the orthogonality relation [3J:
I
f
P (s)P (s)ds -1 m n 2 =f2m+1 ' m = n ~O ,m#nIn the following we will also have use of the recurrence relations:
o ,
(I-s 2)p'(s) + nsP (s) - nP I(s) n n n-o .
(II-6) (II-7) (II-8) (II-9)Now inserting the expansions (6) into (5) and applying the orthogonality relation (7) we arrive at an infinite set of equations:
with 00
L
{a
f'(r) mn n 0=0 I a = (m+j)f
mn -I +!
b f (r)} + f (r)=!
g (r) rmnn m fro sP (s)P (s)ds m n m,n=
0,1, .•..• b mn I -(m+j)f
(I-s )P (s)P'(s)ds 2 m n -I (II-IO) (II-II) (II-12)Using the recurrence relations (8) and (9) we can evaluate the integrals as m+1 = m+1 z;;;+3
•
n = m = m-I a 2m-I•
n mn (II-I3) 0 n " m+1 (m+l) (m+2) n m+1 2m+3b = m(m-I) n m-I (II-14)
mn 2m I
0
n " m+1
To solve the set of equations one has to truncate it after N terms. Then it
can be written in vector notation as:
I
A!' (r) + (I+-B)f(r) r
-I
=
r
~(r) (II-IS)where the vectors
i
and ~ are composed of the functions (fO.fl •••• fN_1) and (gO.gl •..• gN-I). respectively. A and B are square matrices of rank N whose elements are given by (13) and (14). Note that. unlike the usual convention. the indices number from zero to N-I. instead of 1 to N. This is done to remain consistent with the numbering of the Legendre polynomials.
The homogeneous part of (15) allows N independent solutions $ -n . (r) which are the column vectors of the fundamental matrix ~(r). If the latter is known the solution of the inhomogeneous equation can be written down as [4]:
= <p(r) [h+..1.. - r r
f
<p-\t)A-I~(t)dt]
r
o
where h has to be found from boundary conditions imposed upon f after a value for ro has been chosen.
(II-16)
The next chapter will be devoted to the calculation of the fundamental matrix ~(r).
III. The homogeneous solution
The homogeneous part of eq. (II-IS) reads:
A~'(r)
+ (I+.!.B) f(r) = 0 .r
-Let us try as a solution:
ar -1 -2 -N
2
(r) = e C~, ~ = (r , r , ••• r ) •C is a NxN matrix.
(III-I)
(III-2)
Substituting this in (1) we find, after rearrangement:
1
(aA+I)C~ = -(AC'-BC)u (III-3)
r
-Here C' is a matrix composed of column vectors £0' 2~1,···N£N_I' where
£i
are the column vectors of C.
If we equate the coefficients of equal powers of r left and right we find
(aA+I)~O = 0 (III-4a)
(aA+I)c 1 = (nA-B)c
-n+ -n (III-4b)
Q = (NA-B)~N (III-4c)
Let us have a look at (4a) first. Clearly c is an eigenvector of -A and
-0
±
is the corresponding eigenvalue. To calculate the eigenvector take the i-th equation of the set:o .
(III-5)I f we put
(III-6)
then eq. (5) transforms into the recurrence relation (11-8) for Legendre polynomials. All equations of (4a) except the last one are now satisfied. To satisfy the last one (i=N-I) we have to put
If this is true then (7) is the eigenvalue equation of (4a). That this is so can be shown by deriving a recurrence relation for the determinant of A. The lower right hand corner of A-AI looks as follows:
,
0
, ,
"
,
,
"
"
"
N-2 N-I , 2N-S -A2N-T
,
,
,
,
N-I0
,
2N-3
-A"
From this we can derive
,
N-I
- - 0
2N-I N-2 (III-B)
where D.(A) is the determinant of A with rank i and A 's an eigenvalue.
,
SubstitutingD.(A)
,
= (_2)i(i!)2 (2i)! P. (A),
we find that (B) is transformed into the recurrence relationship for Legendre polynomials. Then putting DN(-l/a) = 0 to satisfy (4a) we arrive at (7). Now, if we take an odd number for N, there is one root equal to
zero, which means a and
2
are undefined. So we have to restrict theanalysis to cases with even N, as was already noted by Hammar [2J. For even N the roots occur in pairs with opposite signs. This will have
important consequences later on.
Our next goal is to solve the other column vectors of C. The general expression for an intermediate line of the set (4b) reads:
1 i+1
a~2' 1-I . I 1- ,n+ I + c· 1,n+ I + a -2 · 1.+ 3c . I 1+ ,n+ I
i(n+I)+i(i-l) (i+I)(n+I)-(i+I)(i+2) (111-9)
. I c. I + 2' 3 c. I
21 1- ,n 1.+ 1+ ,n
Now suppose that c· has the form:
1ll C. 1ll Pin - P.(-l/a). n , a (III-IO)
Substituting this in (9) and applying the recurrence relation (II-B) to eliminate the middle term on the left we arrive at:
-2~
~+ I P . 1,n+ I)P, I<-I/a)+1-= i(n+I)p. P. (-I/a) + (i+I)~n-i-I)p. P. (-I/a).
2i-I 1-I,n 1-1 21+3 1+I,n 1+1 (III-II)
Equating the coefficients of P. I and P. I' respectively, we arrive after a
1- 1+ little manipulation at or, p. 1,n+ I Pin 2i+3 i+I+n 2i+I
T+i=ii
(n-i) (n+i+l) 2n+2 with p' = I : 00 p .• 1J 2i+I (-2)jo
(i+j)! (i-j)!j! (III-I2a) (III-I2b) j ~ i (III-I3) J > iThe last step is to check whether eqs. (4c) are satisfied too. The i-th
member of this set reads:
r.
iN i(i-I))\2i-I + 2i I ci-I,N-I (
i+I)N
+ 2' 1+3 (i+I) 2i+3 (i+2)~ )Ci+I,N-I =
°
•Now, because of the triangular form of C, only c
N_I
,
N-I is different from zero in the last column vector. This means that only the equation for whichi = N-2 is not identically zero. But for i = N-2 the coefficient of c
N-I,N-I
vanishes so that this equation is satisfied too.
Since for even N we have N different roots a, all of finite value, we also
have N solutions to the homogeneous equation which can be written as: (l.r
$. = e J C(a.)u, j = O,I, ... N-I •
-J J - (III-I4)
These can be considered to be the column vectors of the fundamental matrix ¢
~
..
lJ n.r i e JI
c. (a.)r -m-I m=o 1m J since c. = 0 for m > i. 1mAlternatively we can write ~ as a power series in r:
4 = I-I 4(m)r-m-IeJ(a)r m=o
The elements of 4(m) are given by: P.(-I/a.) 1 J m a. J (III-IS) (III-16) (III-I7)
The form in which the solution is given here has the disadvantage of being unbounded as r approaches zero. This problem has already been mentioned in the introduction and can lead to difficulties in the calculation of the
inhomogeneous solution. We can, however, convert eq. (15) into a form which
is more suitable for representation at r = O. To this end develop the
exponential in powers of r: The with k i ~ (a.r) ~
..
=I
JI
lJ k! k=o m=o = 2i+1 P. (-I/a.) r 1 J 2i+1 (_2)mI
p=-I (i+m)! (i-m) !m! g. (a.r)p lP Jcoefficients gip have the form:
i (_2)-m (i +m) ! gip
I
(p+m) ! (i-m)!m! m=s s = -p p $ 0 0,
p ~ 0 P. (-I/a.) 1 J m m+1 a. r J (111-18) (111-19)For p s 0 g. can be represented either a
Geg~~bauer
polynomial c(a)(r):n
as a Jacobi polynomial p(a,B)(r)
n
or as
gip
=
(-2)P (i-I2)! 1..
,
p~-p,-p)(O) 1+P gip=
(-2)P (-2/2)! c~-p+j) (-p)! 1+P (0)The value of g. 1p can now easily be determined [3]:
gip 0 i+p odd
Hi?-
(i-I2)! i+p evengip 2i
(i?)!(i?)!
For p ~ 0 we find by inspection:
(l2+i-I)(p+i-3)"'(I2-i+l) t goP
gip
=
(p+i)! lip!Evidently this is zero when i+p odd and p < i.
(III-20a)
(III-20b)
(III-2Ia)
(III-2Ib)
IV. Solution of the inhomogeneous equation
The formal solution of the inhomogeneous equation (II-IS) was already
given in (11-16) and is repeated here for convenience at the same time
introducing a new matrix ~(r):
~(r) <I>(r) [
!:+r'!
I r'l'(t)~(t)dtJ
(IV-I)0
with
'l'(r) <1>-1 (r)A- 1 (IV-2)
Instead of going through the tedious job of inverting <I> and A we shall derive a differential equation for ~. Of course <I>(r) satisfies eq. (III-I):
O. (IV-3)
-I
Multiplying this equation from the right by <I> (r) and applying the rule for differentiation of a product we find
d<l>-I(r)
!
B -A<I> + I + = O.dr r (IV-4)
-I
MUltiplying this from the right by A and from the left by 'l'(r) we obtain
Taking the transpose and mUltiplying by AT from the left we finally
arrive at! T d~T -A - - + dr O. (IV-5) (IV-6)
This equation has the same form as (IV-3) so we may expect a solution similar to the one obtained for ~. It turns out, however, that higher powers of rare needed so let us put
(Iv-7)
Analogously to the previous chapter we now have: T (I-S.A)d I J -n+
o
= 0 T T {(m-n)A +B }d -n T T {(m-N)A +B }!!N-I (IV-Sa) (IV-Sb) (IV-8c)The solution of these equations proceeds along the same lines as in the previous chapter. From (Sa) we have the i-th member:
S i d d - S ..i:2..d = 0
- j 2i+1 i-I,I + il j 2i+1 i+I,1 (IV-9)
which is converted into the recurrence relationship for Legendre polynomials by putting
=
p.O/S.). 1 JFrom (8b) we have:
D i d + d _ D i+1 d =
-fJj
2I'+T
i-l,n+1 i,n+l I-lj 2i+I i+l,n+)i(i+n+2-m) (i+l) (i-n+m-I)
. d. + • d. I 2~+1 ~-1 ,n 21+1 1+ ,n in which we substitute: d. l n P. (I/S.) 1 J S~ J
Following the same procedure as previously we obtain i+n-m+3 i-n+m 1 (i+n-m+3) (i-n+m-2) 2 (n-m+3) (IV-IO) (IV-II) (IV-12) (IV-13a) (IV-I3b)
Noweq. (l3a) is consistent with (10) for n= 0 only i f we choose m= 2. Likewise, if we look at eqs. (Sc) we find that the only member that is not
o
(N-I)(-N+m+N-2) d 2N-3 N-I,N-Iand this too is satisfied by m= 2.
The general expression for q .. now becomes, with q
=
I:1J 00
q .. = (-2)-j (i+j)!
1J (i-j)!j! (IV-14)
T
Since the S. are reciprocal eigenvalues of A they must satisfy the same
J
characteristic equation as the n ..
J
The row vectors of ~ can now be written as:
(IV-IS)
Each of them may still be multiplied by an arbitrary constant. Then ~ has the form:
'I' =
ko!lo (0)
: (N-l)
kN_I!Io
The constants k. have to be found from the identity:
J 'l'A<I>= 1.
The diagonal members of this set of equations yield:
k. !Io(i). (M).= I
1 - - 1
(IV-16)
(IV-I7)
(IV-18)
where, as before, a high index indicates a row vector and a low index a column vector.
Using eqs. (14), (II-I3) and (III-14) this can be evaluated as:
I (IV-19)
To get rid of the exponentials we must arrange the !Io(i) in (16) corresponding to the arrangement of the ~. in ~ such that
-1
S. = -a·.
Then the constant term in (IV-19) yields: N-I
\'
{m
m+
I }k'L d (-a.) - 2 IC I (a.) + 2 +3c I (a.)
~ mo 1 m- m- ,0 1 m m+ ,0 1 m=o
(IV-21)
which, substituting (IV-12) and (111-10) and using the recurrence
relation-ship, becomes: I - = k. 1 N-I
I
(IV-22) m=oUsing the Christoffel-Darboux formula [3] this can further be simplified to:
2 I N ai 2 k i = -2-(a"";:;"'--I-) PN- I (I f a i )· 1 (IV-23)
For r close to zero a similar derivation as the one at the end of part III can be made which yields for the elements of ~:
ljJ .. (r) J1 B.r i = e J
I
m=o I-m d. (S.)r 1m J where g. is given by (111-20). 1P = rP. (If
B.)I
1 J p=-i g. (s.r)p 1p J (IV-24)V. Boundary conditions
The solution for the inhomogeneous equation was given in the previous
chapter as
~(r) (V-I)
It would be convenient if we could choose r = O. To
o do this we first
have to establish that the integrals remain bounded when r approaches o
zero. Now the integrand with number i can be written as: S.t N-I 1 \" = k.e L 1 j=o g. (t) J
j
-k+11.
d·k(S.)t k=o J 1This remains bounded for t + 0 if for every gj:
. I
lim g.(t) < O(tJ - ).
t-+o J
*
(V-2)
It is not difficult to show that when g(k,8) is a smooth function around k = 0 the expansion in Legendre polynomials yields
in fact
~(i)(t).g(t)
is linear in t for t + O. Itg.(t) = O(tj) so that
J
then follows that
~(o) lim<l>(r) ~
r+o
.
.
By the same reason1ng we can make plausible order r1 for small r so let us put as a set
~(o) = (f ,0,0, ••• ). 00
Substituting <I>(r) from (III-IS) we obtain:
j
\" -k-I
(V-3)
that f.(r) must be of the
1 of boundary conditions: (V-4) N-I f. (0) = lim
l:
J r-+o i=o a.r h.e 1 1 k=o L c·k(a.)r J 1 = N-I = liml:
r+o i=o ro h.l:
1 p=-j-I where q -p-I=
0 p s -I p;, -I. *See Appendix A j rPl:
k=q c·k(a.) J 1 p+k+1 a.. . . 1 (p+k+I)! (V-5)If we now select for every j the coefficient of the highest negative power of r we find N-I
I
i=o h. c .. (a.) 1 JJ 1o ,
j O,I, •.• N-l. (V-6)Now c .. (a.) is an even function of a. and if we number the a. such that
J J 1 1 1
N
, n = 0, I, •. "2-1 (V-7)
a2n+1
it is easy to show that the determinant of (6) is zero. In fact, the whole set of equations is satisfied if we put
h
2n+1 -h 2n
N
n = 0,1' ..
0'2-1
(V-B)This leaves us with the task of determining the h2n• To this end observe that, because of the factor exp(a.r), the f2n would explode at infinity
unless:
(i~~ri'
(V-9)
which gives us the desired equations for h
2n• This result allows us to rewrite eq. (I) in a simple and elegant form. Because of the ordering of a
i and eq. (IV-23) we have k2n+1
=
-k2n• If we further define g.(-t)=
(-)Jg.(t) J J (V-IO) we can derive o r 0f
k2n+1!I:
(2n+ I ) (t) .~(t)dt =f
k2n!l: (2n) (t).!!(t)dt -r Instead of (I) we now can write!:
(r) ¢(r)~(r) (V-II) with k ~ 2nf
(2n)- - ! I :
(t+r) .g(t+r)dt r o - (V-12a)00 £Zn+l(r) =
f
f
kZn -r k 00~(Zn)(t).~(t)dt
=
~n
f
~(2n)(t-r).~(t-r)dt.
o (V-12b)We still have to show that the whole of eq. (5) satisfies (4). To this end we use the expressions derived at the end of part III. We have for
f.(r): 1 f. (r) 1 N-I N/2-1
L
j=o ~ .. (r)£.(r) 1J J =L
{~. ~, Z (r)£Z (r) n n n=o N/Z-I ~ 00=fL
{-~i,2n(r)+¢i,Zn+l(r)}
f
n=o rSubstituting (111-18) this becomes, with a
Zn+1 f. (r) 1 N/2-1 [ . . 00 =
1..
~ .~{(_)1+lp.(I/a) ~r
L r 1 Zn L • n=oP=-1
+ P.(I/a Z )L
1 n p=-i + ~. 2 + I (x)f
r k2 ~ (2n) (t) .g(t)dt ] 1 , n -r n -+ ~. Z l(i)£Z I(r)} 1, n+ n+ (V-I3) (V-14)The two terms in curly brackets cancel whenever i+p is even. But when i+p is odd g. is zero for p < i so that this part of f: (r) yields only terms of
-- 1P i 1
the order r and higher. So only for i = 0,1 may we conclude -directly that fi(r) is of the order rio For higher values of i the second part
contains terms of lower order than rio It turns out that these terms
cancel by the summations over m and n but a general proof of this becomes quite complicated and will not be attempted here.
In the numerical calculation of f this effect leads to instabilities at small r and precautions have to be taken to prevent this.
Appendix A Proof that f (r) is of the order rm. m
In Chapter II we used the expansion
f(r,cos8)
m=o
f (r)P (cosS).
m m (A-I)
On the other hand, when we use Cartesian coordinates x=rcos8, y=rsin8, we can, when f is a smooth function of x and y, write it as a power
series in these variables. Since f is even in 8, it is also even in y so that we can write:
f(x,y) = (A-2)
which, converting back to rand 8, reads
fer,S) 00
I
r n C}nJL
ank(l-cos 8) (cos8) 2 k n-2kn=O k=O
(A-3)
Comparing this with (A-I) and applying the ·orthogonality relationship for Legendre polynomials we find a series expansion for f (r):
m
£ (r)
m
The integral can be worked out as follows:
I JPm(U)(I-U2)kun-2kdU -I I = (_)k
Jo
(-)p[~)
f
Pm(U)Un-2PdU -I kL
p=OThis is a well-known integral[6J. Its value 15 (putting n-2p=s)
2s! , s ~ m and m+s even; (5 m)!!(s+m+I)!!
a
s < m or m+s odd. (A-4 ) (A-5) (A-6) The consequence of this lS that the coefficient of rn in (A-4) vanishesfor n < m and also for n > m and n+m even. This then proves that, under the assumption of f being smooth around the origin, f (r) is of the
m
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DETECTION OF TRENDS IN LONG TERM RECORDINGS OF CARDIOVASCUlAR SIGNALS.
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