A general recursive method
by
Ezio Marchi*) and Leopoldo Millán**)
1. Introduction
In this paper we consider a general problem of a recursive method for obtaining
the unique solution described in a recursive way.
Such a method seems that it might be powerful for obtaining solutions in several
branches of physics, chemistry, etc, where complex differential equations, differentialdifference equations, integral differential equations appear.
Before to present it we are going to describe an example where such a general
method appears naturally.
Indeed such a method was already used by Marchi and Millán in an intent to
solve the Hodgkin Huxley differential equation in [2] and with Crinó in solving three
level laser equations in [3].
In the next paragraph we are going to present a simple example where such a
method appears naturally. The next section is related with the solution of the general
method and its validity. Finally, some comments related with the possibility of the
applicability of such a techniques are presented.
*)
Universidad Nacional de San Luis. Instituto de Matemática Aplicada. San Luis-CONICET. Argentina.
Visiting the Department of Applied Mathematics and Alalysis. University of Barcelona, Barcelona,
Spain.
**)
Industrias Pescarmona, Mendoza, Argentina.
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2. A simple example
As we mentioned in the Introduction consider a birth and death stochastic
problem with a finite set of states i  0, 1, , N where the transition probabilities Pi j ( t )
at time t satisfy
Pi, i1 (h)   i h  0(h)
as h  0
i0
Pi, i1 (h)   i h  0(h)
as h  0
i2
Pi, i (h )  1  ( i   i ) h  0(h )
( 2.1)
as i  1
where the parameter take arbitrary values.
i ,  i  0
i  1, 2, 
where 0(h ) in each case may depend on i. The probability for the first state:
P0 ( t )  f (c, t )
( 2 .2 )
is known a priori and it depends on a parameter c. This is equivalent to the case that the
probability for the first state is derived from a differential equation independent from
the following states i  1, 2 ,, n . Thus both situations are possibly to be treated
analogously and therefore we assume that such a probability is computable or known.
The parameter  i and  i are respectively the birth and the death rates and are
arbitrary.
Therefore if Pi ( t ) is the probability to find the system in the state i at time t we
have that since it is a birth and death process; it must satisfy
dPi
  i1 Pi1 ( t )  ( i   i ) Pi (t )   i1 Pi1 (t )
dt
i 1
( 2.3)
i 1
( 2 .4 )
Taking Laplace transform

n i (s)   e s t Pi (t ) dt
0
in the equation (2.3) one obtains
n i  Pi (0)  i1 n i1  ( i  i ) n i   i1 n i1
For more detail we refer to Karlin [1].
Now it is clear that such an expression is a particular case of the general
recurrence equation
2
n k1  a kk1 n k  a kk 11 n k 1    a 0k 1 n 0  b k 1
k  0, 1, 
( 2.5)
where indeed in our particular case the corresponding a’s and b’s can be easily
computed.
Therefore as we have mentioned in the Introduction our task is to solve in
general our recursive problem given by the equation (2.5).
3. The general method
First of all we would like to show that in order to solve the general difference
scheme given in (2.5) it would be of interest to need the first terms or the first relations
which are given by
n 1  a 10 n 0  b1
n 2  (a 12 a 10  a 02 ) n 0  a 12 b1  b 2
n 3  (a 32 a 12 a 10  a 32 a 02  a 13 a 10  a 30 ) n 0
(3.1)
 (a 32 a 12  a 13 ) b1  a 32 b 2  b 3
The good observer will realize that any element of the sum of n 0 is obtained by giving
“jumps” from 3 to 0 and then multiplying the corresponding elements. For example
3  1, 1  0 determines the element a13 a10 . The coefficients of b i , follow a similar
law of formation. Thus, it is important to study the set of possible “jumps”.
Let for given k and r  k the set

E kr  11 , , 1r  :

r

i 1

1i  k, 1i  0
(3.2)
and for an element 11 , , 1r k  E kr define
ak
l1 ,,lr k
 a kk l1 a kk l1l1 l2  a kk ll11 ll22  l3   a kk kl1 lr 1 
with
r  k  k and k  11    1r .
(3.3)
Then, we now will show that
 k 1
k

k

n r  
a
n

k
 l ,,l  0 
 r 1 l1 ,,lr k E kr 1 r 
 i 1
 k 1
 
k

 b i   b k (3.4)
a
k

i

l ,,l 
 r 1 l1 ,,lr k iE kr 1 1 r  
3
is indeed the solution of the difference recurrence equations given in (2.5). In order to
do this, we apply the induction principle. For k=1, we indeed have the first equality of
(3.1). Now suppose that (3.4) it is valid for an arbitrary k, therefore replacing the n i in
(2.5) for i  k , we have
n k 1  a kk 1 n k  a kk 11 n k1    n 0k 1 n 0  b k 1
 k

k 1  k i


k
k



 a kk 1 
a
n

a
b

b
 l ,,l k 0  

i
k
l1 ,,lr k i

i 1  r 1 l1 ,,l r k 1E kr  i
 r 1 l1 ,,lr k E kr 1 r 

 k 1
k  2  k i 1


k 1
k


 a kk 11 
a
n

a
k 1 k 1 l1,,lr k 1  0   
 k i1 l1 ,,lr k i1  b i
k   i 1

r

1
i

1
r

1

l1 ,,lr  E r
l1 ,,lr 
E r



 b k 1 





a   n

 a k2 1 a 2l ,l 2  a 2l 2 n 0  a 2l 1 b1  b 2
1 2
1
1
 a k2 1
2
l1
1
 b1
0

(3.5)
 a 0k 1 n 0  b k 1
We have to show that the sum of all the coefficients of n 0 in (3.5) multiplied by
the corresponding a, equals the coefficients of n 0 in n k 1 . The same has to hold for all
the coefficients of b i .
In order to prove it, we decompose the E kr 1 set a follows:
E kr 1 
k  2 r

l1 1
l , E
1

k 1l1
r i
(3.6)
Since l 2 , l 3    l r  k  1  l i , with r  2 and
l , E
1
k 1l1
r 1
  l ,, l  : l , l ,, l  E
1
r
2
3
r
k 1l1
r 1

(3.7)
using this decomposition, we have
k 1

r 1

a kl,1,l k 1  
1
r
k 1
l1 ,,lr k 1E r
k  2 r

r  2 l1 1
k k  2 r

 a kk 11l
l 1 r  2

l1 ,,lr k 1 l1 E kr 11 l1

l1 ,,lr 1 k 1 lE kr 11l
a kk 11l1 a k 1l1
l1 ,,lr 1 k 1 l1
a kl,1l,l k 1l  a 0k 1
1
r
 a 0k 1
(3.8)
4
k
k 1r

k 1l
  a 0k 1
  a kk 11l  
a
k

1

l


l1 ,,l r 
 r 1 l1 ,,lr k 1lE kr 1l

l1
Where from the first to the second expressions we applied a change of indeces in
the sum, namely:
2  r  k  1, 1  1  k  2  r to 1  1  k, 2  r  k  2  1 .
The third expression is obtained by the change of variable r to r-1. The last term in (3.8)
is indeed the same as the sum of all the coefficients of n 0 in (3.5). Now study the other
terms.
For the b i , we have
k

l1
k 1i

 r 2
k  2i r

l1

l1 ,,lr 1 k 1i lE kr 11i l
k 1  k i
   a kk 11l
l1 
 l1
k  2 i  l
k 1  k i
   a kk 11l
l1 
 l1
k 1i l
k 1
 k l
  a kk 11l 
 i1
l1
k 1i l

r 2

r 1

r 1
 k
a kk 11l a kl,1l,l k 1i l b i    a ik b i  b k 1
1
r 1
 i1

l1 ,,lr 1 k 1i lE kr 11i l
 k
a kl,1l,l k 1i l b i    a ik 1 b i  b k 1 (3.10)
1
r 1
 i1

k

a kl,1l,l k 1i l  b i   a ik 1 b k  b k 1
1
r

i 1

k

a kl,1l,l k 1i l  b i   a ik 1 b i  b k 1
1
r

i 1
l1 ,,lr k 1i lE kr 1i l
l1 ,,lr k 1i lE kr 1i l
where the first equality is obtained by just to reorder terms. The second one is obtained
by change of indeces: 2  r  k  1  i  1  i and 1  1  k  2  i  r to 1  1  k  i and
2  r  k  2  i  1 . The third expression is derived by the change of variable r to r-1
and finally by changing 1  i  k  1 and 1  1  k  i to 1  1  k  1 and 1  i  k  1
the last equality holds. The last term in (3.10) is indeed all the sum of all remaining
parts in the expression of n k 1 without the factor n 0 . Thus, we have completed the
proof.
4. Concluding remarks
We would like to mention as concluding remarks that the formula obtained in
the previous paragraph is rather powerful for studying problems already mentioned in
5
the Introduction. Moreover it seems that an conceptual point of view that it perhaps it
might be useful for attacking from a different point of view the functional integrals. We
hope to develop it in the future.
References
(1) Karlin, S. & Taylor: First course of Stochastic Processes. Vol. I. Academic
Press.
(2) (2) Marchi, E. and Millán, L.: On the exact solution of differential equations of
Hodgkin-Huxley model type. Proceeding of the first International Conference on
Mathematical Modeling. Vol. 2, 447-458 (1977).
(3) Marchi, E., Millán, L. and Crinó, E.: Exact solution of the rate equations for the
three energy level lase. Quantum Electronics. IEFE. Vol. OE-16, 4, 461-464
(1980).
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