Open Journal of Discrete Mathematics, 2011, 1, 66-84
doi:10.4236/ojdm.2011.12009 Published Online July 2011 (http://www.SciRP.org/journal/ojdm)
Mean Square Numerical Methods for Initial Value
Random Differential Equations
1
Magdy A. El-Tawil1*, Mohammed A. Sohaly2
Engineering Mathematics Department, Faculty of Engineering, Cairo University, Giza, Egypt
2
Mathematic Department, Faculty of Science, Mansoura University, Mansoura, Egypt
E-mail: [email protected], [email protected]
Received April 1, 2011, revised April 30, 2011, accepted May 9, 2011
Abstract
In this paper, the random Euler and random Runge-Kutta of the second order methods are used in solving
random differential initial value problems of first order. The conditions of the mean square convergence of
the numerical solutions are studied. The statistical properties of the numerical solutions are computed
through numerical case studies.
Keywords: Random Differential Equations, Mean Square Sense, Second Random Variable, Initial Value
Problems, Random Euler Method, Random Runge Kutta-2 Method
1. Introduction
Random differential equations (RDE) are defined as differential equations involving random inputs. In recent
years, increasing interest in the numerical solution of
(RDE) has led to the progressive development of several
numerical methods. This paper is interested in studying
the following random differential initial value problem
(RIVP) of the form:
dX
 f  t , X  , X  t0   X 0
dt
(1.1)
Randomness may exist in the initial value or in the differential operator or both. In [1,2], the authors discussed
the general order conditions and a global convergence
proof is given for stochastic Runge-Kutta methods applied to stochastic ordinary differential equations
(SODEs) of Stratonovich type. In [3,4], the authors discussed the random Euler method and the conditions for
the mean square convergence of this problem. In [5], the
authors considered a very simple adaptive algorithm
based on controlling only the drift component of a time
step. Platen, E. [6] discussed discrete time strong and
weak approximation methods that are suitable for different applications. Other numerical methods are discussed
in [7-12].
In this paper the random Euler and random RungeKutta of the second order methods are used to obtain an
approximate solution for Equation (1.1). This paper is
Copyright © 2011 SciRes.
organized as follows. In Section 2, some important preliminaries are discussed. In Section 3, the existence and
uniqueness of the solution of random differential initial
value problem is discussed and the convergence of random Euler and random Runge-Kutta of the second order
methods is discussed. In Section 4, the statistical properties for the exact and numerical solutions are studied.
Section 5 presents the solution of some numerical examples of first order random differential equations using
random Euler and random Runge-Kutta of the second
order methods showing the convergence of the numerical
solutions to the exact ones (if possible). The general conclusions are presented in the end section.
2. Preliminaries
2.1. Mean Square Calculus [13]
Definition1. Let us consider the properties of a class of
real r.v.’s X 1 , X 2 , , X n whose second moments,
E X 12 , E X 22 , are finite. In this case, they are called
   
“second order random variables”, (2.r.v’s).
Definition 2. The linear vector space of second order
random variables with inner product, norm and distance,
is called an L2 -space.
A s.p.  X  t  , t  T  is called a “second order stochastic process” (2.s.p) if for t1 , t2 , tn , the r.v’s
 X  t1  , X  t2  ,, X  tn  are elements of L2 -space.
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M. A. EL-TAWIL ET AL.
A second order s.p.
 X t  , t  T 
  t    ,
X t   E X
2
2
is characterized by
t T .
2.1.1. The Convergence in Mean Square
A sequence of r.v’s  X n  converges in mean square
(m.s) to a random variable X
m. s
X or
if lim X n  X  0 i.e. X n 
n 
lim X n  X
n 
where lim is the limit in mean square sense.
2.1.2. Mean-Square Differentiability
The random process  X  t  is mean-square differentiX  Xt
exists, and is denoted by
able at t if lim t  h
h 0
h
X  Xt
 X t
lim t  h
h 0
h
suppose the right-hand side function b  t , X  is continuous and satisfies a mean square (m.s) Lipschitz condition in its second argument:
b t, X   b t, Y   c X  Y
where C is a constant or
b t, X   b t, Y   c t  X  Y  c X  Y
(3.4)
where c(t) is a continuous function {because in every
finite interval c(t) ≤ constant}.
then the solution of Equation (3.1) exists and is unique.
The proof
The existence can be proved by using successive approximations. Let
X t0  X 0
(3.5)
and for n  1

t

X tn  X 0  t 0 b X sn 1 , s ds.
X t   X t
1
3.1. Existence and Uniqueness
(3.6)
0
t 0 b  s, X 0  ds
t

 k  t  t0
where
Let us have the random initial value problem
dX
 b  t , X  , t  T   t0 , t  , X  t0   X 0
dt
b t, X   k
(3.1)
where X  t  is second order random process. This
equation is equivalent to integral equation
X  t   X 0  t 0 b  X  s  , s  ds
t
(3.7)
For n > 1 we obtain:
X t   X t
n
n 1
(3.2)
Theorem (3.1.1)
If we have the random initial value problem (3.1) and
n
(3.3)
For n  1 we obtain:
3. Random Initial Value Problem (RIVP)
X t   X t
67

n 1
n2
t 0 b  s, X s   b  s, X s  ds
t
 n 1
t
 t 0 c  X s
 n  2
 Xs
(3.8)
ds
Successively, we can obtain the following:
t
n 1
  c X sn 1  X sn  2 ds
t0
t
t
t0
t0
t t t
  c  c X sn  2  X sn 3 dsds  c3    X sn 3  X sn  4 dsdsds
t 0 t 0 t0
t t
t
t t
t
t0 t0
t0
t0 t 0
t0
 c n 1    X s1  X s0 ds  ds  c n 1    k s  t0 d  ds.
Hence
X X
n
n 1
 kc
n 1
t t t


t  t0

d
]d

d
s
t
s
s
s


 ds  kc n 1


0
   
n!


t0  t0
t0 
 t0
t
n
(3.9)
Since:

 k .c
n 1
n 1
t  t0
n!

k c t  t0
e
c
is convergent for finite t,
(3.10)
hence we can have the following
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M. A. EL-TAWIL ET AL.
68
 k t  t0   kc t  t0
X t1  X t0  X t2  X t1    X tn 1  X tn  2    

2!
 1!  
2
n


 kc n  2 t  t0 n 1 
t  t0
   
     k .c ( n 1)


n!
 n  1! 
n 1


(3.11)
Accordingly,
X t1  X t0  X t2  X t1    X tn 1  X tn  2  X tn  X tn 1  lim X tn  X t0  X t
n 
Hence:
X t1  X t0  X t2  X t1   X tn 1  X tn  2  X tn  X tn 1  X t1  X t0  X t2  X t1    X tn 1  X tn  2  X n  X n 1 
This yield lim X tn  X t0  X t 
n 
constant, hence the only solution of the integral Equation
(3.17) is
k c t  t0
e
c
Then lim X n exists. i.e.
Ut  0
n 
X t  lim X
n 
n
t
(3.12)
Since X tn is the general solution of Equation (3.6)
and X t is the general solution of Equation (3.2).
To prove the uniqueness of the solution, let X t is a
solution of the initial-value problem (3.1), or, which is
the same, of the integral Equation (3.2), and Yt is the
solution of
dy
 b Y  t  , t  , t  T  t0 , t  , Y  t0   Y0
dt
(3.13)
to prove the uniqueness of the solution we want to prove
that
X t  Yt .
(3.14)
By subtraction (3.2) and the corresponding integral
equation for Yt
t
3.2. The Convergence of Euler Scheme for
Random Differential Equations in
(m.s.) Sense
Let us have the random differential equation
X  t   f  X  t  , t  , t  T  t0 , t1  , X  t0   X 0 (3.22)
where X0 is a random variable and the unknown X  t 
as well as the right-hand side f (X,t) are stochastic processes defined on the same probability space.
Definitions [6,7]
 Let g: T  L2 is an m.s. bounded function and let h
> 0 then
The “m.s. modulus of continuity of g” is the function
W  g , h   sup
t t*  h
t0
(3.15)
t
X t  Yt  t c  X s  Ys ds
0
t
(3.16)
i.e; U  t 0 c  U s ds
(3.17)
where U t  X t  Yt .
From Equation (3.17) we have:
(3.18)
t
U t  c t  t0 U t
(3.19)
Note that: at t  t0 we obtain U 0  0 then:
U0  0
From (3.19) c must satisfy the following condition:
c
1
t  t0
(3.20)
which is in contradiction with being an independent free
Copyright © 2011 SciRes.
(3.21)
Hence X t  Yt i.e., the solution of Equation (3.1) exists and is unique.
X t  Yt  X 0  Y0    b  X s , s   b Ys , s  ds
Since X 0  Y0 then:
k c t  t0
e
c
 
g t   g t* , t, t*  T
 The function g is said to be m.s uniformly continuous
in T if:
lim W  g , h   0
h 0
Note that:
(The limit depends on h because g is defined at every t
so we can write W  g , h   W  h  )
In the problem (3.22), we find that the convergence
of this problem depends on the right hand side (i.e.
f  X  t  , t  then we want to apply the previous definition on f  X  t  , t  hence:
Let f  X  t  , t  be defined on S  T where S is
bounded set in L2
Then we say that f is “randomly bounded uniformly
continuous” in S, if
lim W  f  x,. , h   0
h 0
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M. A. EL-TAWIL ET AL.


(note that W f  X . , h   W  h  )

E

3.2.1. Random Mean Value Theorem for Stochastic
Processes
The aim of this section is to establish a relationship between the increment X  t   X  t0  of a 2-s.p. and its
m.s. derivative X   for some  lying in the interval
t0 , t  for t  t0 . This result will be used in the next
section to prove the convergence of the random Euler
method.
Lemma (3.3.2) [6,7]
Let Y  t  is a 2-s.p., m.s. continuous on interval
T  t0 ,t  . Then, there exists   t0 , t  such that
t
t
0
Y  s ds  Y    t  t0  , t0  t1  t
(3.25)
The proof
Since Y  t  is m.s. continuous, the integral process
t
t
0
Y  s ds is well defined and the correlation function
 y  r , s  is well defined, is a deterministic continuous
function on T × T.
For each fixed r, the function  y  r , s  is continuous
and by the classic mean value theorem for integrals, it
follows that:
 y  r , s  ds   y  r ,    t  t 0 
t
t
0
  t0 , t 
t
0
(3.26)
E  y  r  y  s   ds  E  y  r  y     t  t0 
Since  y  r , s   E  y  r  y  s  
We must prove that for the value  satisfying (3.26)
one get:
t y  s  ds  y    t  t0 
t
2

t
t0

2

y  s  ds  y    t  t0    0

  y  s  ds  y   t  t  


 E   y  s  ds    2 E   y  s  ds  y     t  t 




2
t
0
t0
t
2
t0
t
t0
0
2
2
 E  y     t  t0  


Copyright © 2011 SciRes.

then by substituting in (3.28)

E


t
t
t

2

y  s  ds  y     t  t0  

t0
 t
t
0
E  y  s  y  r  drds  t E  y  s  y   ds   t  t0 
0
t
0
 t E  y  s  y   ds   t  t0   E[ y   y    t  t0 
t
2
0
(3.29)
And since:
t
t
0
E  y  r  y  s   d s  E  y  r  y     t  t0 
then by substituting in (3.28) we have:

E



2

y  s  ds  y    t  t0  

t
t0
  E  y  s  y   ds   t  t0    E  y  s  y   ds   t  t0 
t0
t0
t
t
 E[ y   y    t  t0   E[ y   y    t  t0   0
2
i.e.
2
t y  s  ds  y    t  t0 
t
0
2
 0 we obtain
t y  s  ds  y    t  t0 
0
Theorem (3.3.1) [6,7]
Let X  s  be a m.s. differentiable 2-s.p. in t0 , t1 
and m.s. continuous in T  t0 , t  . Then, there exists
  t0 , t1  such that X  t   X  t0   X    t  t0  ,
t0  t1  t
The proof
The result is a direct consequence of Lemma (3.3.2)
applied to the 2-s.p. Y  t   X  t 
X  s  ds X    t  t0 
t
t
0
t
t
0
X  s  ds  X  t   X  t0 
(3.30)
The proof of (3.30)
Let X  t  be a m.s. differentiable on T and let the ordinary function f  t , s  be continuous on T  T whose
f  t , s 
partial derivative
exist
s
t
If Y  t   f  t , s  X  s  ds
(3.31)
a
Then
Y  t   f  t , s  X  s  a  a
t
(3.28)
and since:
t0
2
t t

y  s  ds     E  y  s  y  r  drds
t0 t0

(3.27)
The proof of (3.27)
As

E

t
And the integral formula
0

 E


t
Note that by definition of  y  r , s  expression (3.26)
can be written in the form
t
69
t
f  t , s 
s
X  s  ds
(3.32)
Let f  t , s   1 in Equations (3.31) and (3.32) we
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Now we have the solution of problem (3.22) is X  tn 
At t  tn then X  tn   X  t  and the solution of
Euler method (3.33) is X n
Then we can define the error
have the useful result that:
If X  t  is m.s. Riemann integrable on T then:
a X  s  ds  X  t   X  a ,  a, t   T
t
en  X n  X  tn 
Then we have:
X  t   X  t0   X    t  t0 
en  X n  X  t 
By (3.33) and (3.36) it follows that
3.2.2. The Convergence of Random Euler Scheme
In this section we are interested in the mean square convergence, in the fixed station sense, of the random Euler
method defined by
X n 1  X n  hf  X n , tn  , X  t0   X 0 , n  0
X n 1  X  tn 1 

This implies
(3.33)


en 1  X n  X  tn   h f  X n , tn   f X  t  , t
where X n and f  X n , tn  are 2-r.v.’s , h  tn  tn 1 ,
tn  t0  nh and f: S  T  L2 , S  L2 satisfies the
following conditions:
C1: f  X , t  is randomly bounded uniformly continuous,
C2: f  X , t  satisfies the m.s. Lipschitz condition

Hence


en 1  X n  X  tn   h f  X n , tn   f X  t  , t



 X n  X  tn   h f X  t  , t  f  X n , tn 
f  x, t   f  y , t   k  t  x  y
(3.37)
Since:
where
t k  t   
t1






Note   t0 , t1  and we can use  instead of t
and from Theorem (3.3.1) at t  t  then we have:
X  t   X  t0   X  t   t  t0  then


X  t   X  t0   f X  t , t   t  t0 
Note that we deal with the interval  tn , tn 1   t
  tn , tn 1  and hence t0 was the starting in the prob-
lem (3.22) and here tn is the starting and since Euler
method deal with solution depend on previous solution
and if we have X  tn  instead of X  t0  then we can
use X  tn 1  instead of X  t  .
Then the final form of the problem (3.22) is
X  tn 1   X  tn   hf X  t , t  , for some

t   tn , tn 1 
Copyright © 2011 SciRes.

(3.36)
 

 f  X  tn  , tn   f  X  tn  , tn   f  X n , tn 
(3.38)
 f ( X (t ), t )  f ( X (t ), tn )
(3.35)
where X  t  is the theoretical solution 2-s.p. of the
problem (3.22), t  tn  t0  nh .
Taking into account (3.22), and Theorem (3.3.1), one
gets,
Since from (3.22) we have at t  t then

X t   f x t  , t
 
 f X  t  , t  f X  t  , tn  f X  t  , tn
Note that under hypothesis C1 and C2, we are interested in the m.s. convergence to zero of the error
en  X n  X  t 

f X  t  , t  f  X n , tn 
(3.34)
0


 X n  hf  X n , tn   X  tn   hf X  t , t 
 f ( X (t ), tn )  f ( X (tn ), tn )
 f ( X (tn ), tn )  f ( X n , tn )
Since the theoretical solution X is m.s. bounded in
t0 , t1  , sup X  t   M   and
t0  t  t1
Under hypothesis C1, C2 We obtain



 
  wh
f  X  t  , t   f  X  t  , t   k  t  Mh
f X  t  , t  f X  t  , tn

n
n
n
n
(*)
Since k  tn  is Lipschitz constant (from C2) and from
Theorem (3.3.1) we have X  t   X  t0   X    t  t0 
and note that the two points are X  t  and X  tn  in
(*) then we have:
X  t   X  tn   X   t  tn  Mh
Since t  tn  h and M  sup X  t 
t0  t  t1

f  X  tn  , tn   f  X n , tn 
 k  tn  X  tn   X n  k  tn  en
OJDM
M. A. EL-TAWIL ET AL.
71
Then by substituting in (3.38) we have


f X  t  , t  f  X n , tn   w  h   k  tn  Mh  k  tn  en
(3.39)
Then by substituting in (3.37) we have
en 1  en  h  w  h   k  tn  Mh  k  tn  en   1  k  tn  h  en  h  w  h   k  tn  Mh 
 1  K  tn  h  [1  k  tn  h  en 1  h  w  h   k  tn  Mh   h  w  h   k  tn  Mh 
 1  K  tn  h  en 1  h  w  h   k  tn  Mh  1  1  k  tn  h  
2
3
2
 1  K  tn  h  en  2  h  w  h   k  tn  Mh  1  1  k  tn  h   1  k  tn  h  


 1  K  tn  h 
n 1
2
n
e0  h  w  h   k  tn  Mh  1  1  k  tn  h   1  k  tn  h     1  k  tn  h  


Since:
1  1  k  t  h   1  k  t  h 2    1  k  t  h n 
n
n
n


is geometrical sequence.
Then:
 w  h   k  tn  Mh 
 0 as
The term: lim 
h 0
k  tn 
h  0 (  w  h   0 as h  0 )
And the second term:
 w  h   k  tn  Mh 
1  K  t  h n 
lim 
n
h 0


k  tn 
1  1  k  t  h   1  k  t  h 2    1  k  t  h n 
n
n
n


1  k  t  h 

n
n
1
we have:
k  tn  h
 w  h   k  tn  Mh 
n
lim 
[1  K  tn  h  ]
h 0
k  tn 
Then we get
en 1  1  K  tn  h 
n 1
e0
[1  k  tn  h   1]
n
  w  h   k  tn  Mh 
k  tn 
Taking into account that e0  0 where
1  k  t  h n  1
n


  w  h   k  tn  Mh  
k  tn 
lim en 1
h 0
1  k  t  h n  1
n


 lim  w  h   k  tn  Mh  
h 0
k  tn 
 w  h   k  tn  Mh 
1  K  t  h n 
 lim 
n


h 0
k  tn 
 w  h   k  tn  Mh 
 lim 
h 0
k  tn 
(3.40)
Note that:
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 w  h   k  tn  Mh 
n
lim 1  K  tn  h  
 lim 

0
h 0
h


k  tn 
(3.42)
The first limit in (3.42) equal zero and:
n
The computation of lim 1  K  tn  h   as follows:
h 0 

n
Let y  lim 1  K  tn  h   then by tacking the loh 0 

garithm of the two sides we have:
e0  X 0  X  t0   0 .
en 1
(3.41)
n
ln y  ln lim 1  K  tn  h  
h 0 

n
 lim ln 1  K  tn  h  
h 0



 lim n ln 1  K  tn  h  
h 0
t t
 lim n 0 ln 1  K  tn  h  
h 0
h
t n  t0
ln 1  K  tn  h  
 lim
h 0
h
 tn  t0  ln 1  k  tn  h 
 lim
h 0
h
By using the (L’Hospital’s Rule):
OJDM
M. A. EL-TAWIL ET AL.
72
lim
 tn  t0  ln 1  k  tn  h 
h 0
problem (3.22), t  tn  t0  nh .
Taking into account (3.22), and Theorem (3.3.1), one
gets,
Since from (3.22) we have at t  t then
h
 t n  t0 
 lim
1
k  tn 
(1  k  tn  h)
1
h 0
 t n  t0  k  t n 
  t  t0  k  t 
h 0 1  k  t  h
n
ln y   t  t0  k  tn  which implies that
y  e
t  t0  k  t n 
X  t   X  t0   X  t   t  t0  

X  t   X  t0   f X  t , t  t  t0 
hence
By substituting in (3.42):
 0  e
t  t0  k  t n 
(3.44)
0
By substituting from (3.44) and (3.42) in (3.40) hence
lim en 1  0 i.e.,
h 0
en  converge in m.s to zero as
h  0 hence X n  X  tn   X  t  .
m. s
3.3. The Convergence of Runge-Kutta of Second
Order Scheme for Random Differential
Equations in Mean Square Sense
In this section we are interested in the mean square convergence, in the fixed station sense, of the random
Runge-Kutta of second order method defined by
h
X n 1  X n  [ f  X n , tn   f  X n  f  X n , tn  , tn 1  ,
2
X  t0   X 0 , n  0
(3.45)
where X n and f  X n , tn  are 2-r.v.’s, h  tn  tn1 ,
tn  t0  nh and f: S  T  L2 , S  L2 satisfies the following conditions:
C1: f  X , t  is randomly bounded uniformly continuous,
C2: f  X , t  satisfies the m.s. Lipschitz condition
f  x, t   f  y , t   k  t  x  y
t1
0
(3.46)
Note that under hypothesis C1 and C2, we are interested in the m.s. convergence to zero of the error
en  X n  X  t 
(3.47)
where X  t  is the theoretical solution 2-s.p. of the
Copyright © 2011 SciRes.
lem (3.22) and here tn is the starting and since Euler
method deal with solution depend on previous solution
and if we have X  tn  instead of X  t0   we can
use X  tn 1  instead of X  t  then the final form of
the problem (3.22) is


X  tn 1   X  tn   hf X  t , t  , for some t   tn , tn 1 
(3.48)
Now we have the solution of problem (3.22) is X  tn 
At t  tn then X  tn   X  t  and the solution of
Runge-Kutta of 2 order method (3.45) is X n
Then we can define the error
en  X n  X  t 
By (3.45) and (3.48) it follows that
X n 1  X  tn 1 
h
 f  X n , tn   f  X n  f  X n , tn  , tn 1    X  tn 

2
h
h
 f X  t  , t  f X  t  , t
2
2
 Xn 




Then we obtain:
en 1  X n  X  tn  




h
f  X n , tn   f X  t  , t
2





h
f  X n  f  X n , tn  , tn 1   f X  t  , t
2
By taking the norm for the two sides:
en 1  X n  X  tn  
where
t k  t   

Note that we deal with the interval  tn , tn 1   t
  tn , tn 1  and hence t0 was the starting in the prob-
t t k t
lim 1  K  tn  h    e 0   n 
h 0 

n
 w  h   k  tn  Mh 
1  K  t  h n 
lim 
n


h 0
k  tn 

Note   t0 , t1  and we can use  instead of t
And from Theorem (3.3.1) at t  t then we obtain
 lim
Then

X  t   f x  t  , t
(3.43)

h
f  X n , tn   f X  t  , t
2





h
f  X n  f  X n , tn  , tn 1   f X  t  , t
2
h
 X n  X  tn   f X  t  , t  f  X n , tn 
2
h
 f  X n  f  X n , tn  , tn 1   f X  t  , t
2
(3.49)



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M. A. EL-TAWIL ET AL.
73
Since:


f X  t  , t  f  X n , tn 

 
 
 f X  t  , t  f X  t  , tn  f X  t  , tn

 f  X  t n  , t n   f  X  t n  , tn   f  X n , t n 

 

(3.50)


 f X  t  , t  f X  t  , tn  f X  t  , tn  f  X  tn  , tn 
 f  X  tn  , tn   f  X n , tn 
Are X  t  and X  tn  in (*) then
Since the theoretical solution X is m.s. bounded in
t0 , t1  , sup X  t   M   and
X  t   X  tn   X   t  tn  Mh
t0  t  t1
Under hypothesis C1, C2 We have



 
  wh
f  X  t  , t   f  X  t  , t   k  t  Mh
f X  t  , t  f X  t  , tn

n
n
n
n
where t  tn  h and M  sup X  t 
t0  t  t1
(*)
f  X  tn  , tn   f  X n , tn 
where k  tn  Is Lipschitz constant (from C2) and:

From Theorem (3.3.1) we have
X  t   X  t0   X    t  t0  and note that the two points
Then by substituting in (3.50) we have

 k  tn  X  tn   X n  k  tn  en

f X  t  , t  f  X n , tn   w  h   k  tn  Mh  k  tn  en
(3.51)
And another term:

f  X n  f  X n , tn  , tn 1   f X  t  , t



f  X t  , t   f  X t  , t   f  X t  , t 
 f X  t  , t  f  X n  f  X n , tn  , tn 1 





n 1
n 1
 f  X  tn  , tn 1   f  X  tn  , tn 1   f  X n  f  X n , tn  , tn 1 

 



 f X  t  , t  f X  t  , tn 1  f X  t  , tn 1  f  X  tn  , tn1 
 f  X  tn  , tn 1   f  X n  f  X n , tn  , tn 1 
 w  h   k  tn 1  Mh  k  tn 1   en  M 
Since:



 

f  X t  , t   f  X t  , t 
f X  t  , t  f X  t  , tn 1  w  h 

n 1
n
n 1
 k  tn 1  Mh
where k  tn  . Is Lipschitz constant (from C2) and:
From Theorem (3.3.1) we have
X  t   X  t0   X    t  t0  and note that the two points
are X  t  and X  tn  in (*) then we have:
X  t   X  tn   X   t  tn  Mh
Copyright © 2011 SciRes.
where t  tn  h and M  sup X  t 
t0  t  t1
And the last term:
f  X  tn  , tn 1   f  X n  f  X n , tn  , tn 1 
 k  tn 1  X  tn   X n  f  X n , tn 
 k  tn 1  X  tn   X n  f  X n , tn 
 k  tn 1   en  M 
Then by substituting in (3.49) we have
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M. A. EL-TAWIL ET AL.
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h
h
 w  h   k  tn  Mh  k  tn  en    w  h   k  tn 1  Mh  k  tn 1   en  M  
2
2
 h
 h
 en 1  k  tn   k  tn 1     2w  h   hk  tn  M  hk  tn 1  M  K  tn 1  M 
2

 2


 h
 h
  en 1 1  k  tn   k  tn 1     2 w  h   hk  tn  M  hk  tn 1  M  K  tn  M  
 2
 2


en 1  en 
 h
 h
1  k  tn   k  tn 1      2w  h   hk  tn  M  hk  tn 1  M  K  tn 1  M 
 2
 2
2
h
 h
 h


 en 1  1  k  tn   k  tn 1     2 w  h   hk  tn  M  hk  tn 1  M  K  tn 1  M   2  k  tn   k  tn 1  
2
 2
 2




 h
 h
  en  2 1  k  tn   k  tn 1     2w  h   hk  tn  M  hk  tn 1  M  K  tn  M  
2
2




2
h
 h
  h


1  k  tn   k  tn 1      2 w  h   hk  tn  M  hk  tn 1  M  K  tn 1  M   2  k  tn   k  tn 1  
2
  2


 2
3
 h
 h
 en  2 1  k  tn   k  tn 1     2w  h   hk  tn  M  hk  tn 1  M  K  tn 1  M 
2

 2
2
  h
  h
 
1  1  k  tn   k  tn 1    1  k  tn   k  tn 1   
  2
 
  2
Then we have:
n 1
h
 h

en 1  e0 1  k  tn   k  tn 1     2 w  h   hk  tn  M  hk  tn 1  M  K  tn1  M 
2
 2

2
n
  h
  h

 h
 











k
t
k
t
k
t
k
t
k
t
k
t
1
1
1
1
 n   n 1   
 n   n 1  
 n   n 1   
 

  2

 2
 
  2
Since:
2
n
  h
  h

 h
 
1  1  k  tn   k  tn 1    1  k  tn   k  tn 1      1  k  tn   k  tn 1   
  2

 2
 
  2
is geometrical sequence then we have:
n
 h
 


1
k
t
k
t





 1


n
n

1
n
2
 
 2
 h
  h

 h

[1   1  k  tn   k  tn 1     1  k  tn   k  tn 1       1  k  tn   k  tn 1   ] 
h
 2
  2

 2

k  tn   k  tn 1 
2
Then we get:
en 1
n
 h
 


k
t
k
t
1





 1



1
n
n
n 1
 
 2
h
 h


 e0 1  k  tn   k  tn 1     2 w  h   hk  tn  M  hk  tn 1  M  K  tn 1  M 
h
2
 2

k  tn   k  tn 1 
2
Taking into account that e0  0 where
e0  X 0  X  t0   0
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75
n
 h
 
1  k  tn   k  tn 1     1
 
 2
h
  2 w  h   hk  tn  M  hk  tn 1  M  k  tn 1  M  
h
2
k  tn   k  tn 1 
2
en 1
n
 h
 
k
t
k
t
1







n
n 1    1
 
 2
h

lim en 1  lim  2 w  h   k  tn  Mh  hk  tn 1  M  K  tn 1  M 
h 0
h 0 2
h
k  tn   k  tn 1 
2
h
n
 2 w  h   k  tn  Mh  k  tn 1  Mh  k  tn 1  M   h
 
K
t
k
t
1
 lim 2







n
n 1  

h 0
h
 
 2
k  tn   k  tn 1 
2
h
 2 w  h   k  tn  Mh  k  tn 1  Mh  k  tn 1  M 
 lim 2
h 0
h
k  tn   K  tn 1 
2
Note that:
The term:
h
 2w  h   k  tn  Mh  k  tn 1  Mh  k  tn 1  M 
0
lim 2

0
h 0
h
k  tn 1 
k  tn   K  tn 1 
2
and the second term:
h
n
 2w  h   k  tn  Mh  k  tn 1  Mh  k  tn 1  M   h
 
lim 2
1  K  tn   k  tn 1   


h 0
h
 
 2
k  tn   k  tn 1 
2
we have:
h
n
 2w  h   k  tn  Mh  k  tn 1  Mh  k  tn 1  M  
 h
 
lim 2
1  K  tn   k  tn 1   


h 0
h
 
 2
k  tn   k  tn 1 
2
h
n
 h
 2 w  h   k  tn  Mh  k  tn 1  Mh  k  tn 1  M 
 
lim  1  K  tn   k  tn 1   
 lim 2
h 0
h 0 
h
2
 

k  tn   k  tn 1 
2
The first limit in (3.53) equals zero and:
The computation of
n
 h
 
lim 1  K  tn   k  tn 1    is as follows:
h 0
 
 2
 h

ln y  ln lim  1  K  tn   k  tn 1  
h 0 
2


(3.53)
n
 h
 
Let y  lim  1  K  tn   k  tn 1    then by tackh 0
 
 2
ing the logarithm of the two sides we have:
n
n

 h
 
ln  1  K  tn   k  tn 1   
  lim
 
 h 0  2
  h
t t
 
 lim n ln 1  K  tn   k  tn 1      lim n 0

0
h 0
h
h
2
 
 
h
 tn  t0  ln(1  k  tn   k  tn 1 
2


 lim
h 0
h
Copyright © 2011 SciRes.
(3.52)
  h

ln[1  2 K  tn   k  tn 1   

 
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76
A sequence of r.v’s  X n  converges in probability to
a random variable X as n   if
By using the (L’Hospital’s Rule):

h

k  tn   k  tn 1   

  2
lim
h 0
h
1
1

k  tn  
 t n  t0 
 h
  2

1  k  tn   k  tn 1  
2


 lim
h 0
1
1
1
 t n  t0  k  t n 
 t  t0  k  t 
 lim 2
 2
h 0
h
1  k t 
1  k  tn   k  tn 1 
2
 tn  t0  ln 1 
 t  t0  k  t 
ln y 
2 1  k  t  
Then y  e
 t  t0  k  t 
2 1 k  t  
lim p  X n  X     0   0
n 
(3.54)
hence:
t t k t 
0
n
 h
 
2 1 k  t  
lim 1  K  tn  h  k  tn 1     e 
h 0
 
 2
By substituting in (3.53):
h
 2w  h   k  tn  Mh  k  tn 1  Mh  k  tn 1  M 
lim 2
h 0
h
k  tn   k  tn 1 
2
t t k t 
0
n
 h
 
2 1 k  t  
0
1  K  tn   k  tn 1     0  e 
 
 2
(3.55)
By substituting from (3.55) and (3.53) in (3.51) then
we obtain lim en 1  0 i.e.
h 0
en 
converges in m.s to
zero as h  0 hence X n  X  tn   X  t 
m. s
4. Some Results
Definition 4.2 [13]. “The convergence in distribution”
A sequence of r.v’s  X n  converge in distribution to
a random variable X as n   if
lim Fxn  x   Fx  x 
n 
Lemma (4.1) [13]
The convergence in m.s implies convergence in probability
Lemma (4.2) [13]
The convergence in probability implies convergence
in distribution
Theorem 4.2
m. s
m. s
X then PDF of  X n  
If X n 
PDF of
f xn  x   f X  x 
 X  i.e.; nlim

Proof
m. s
X then
Since we have shown that If X n 
d
X n 
X
m. s
X then lim Fxn  x   FX  x 
i.e., if X n 
n 
d
d
Then we have: lim FX n  x  
FX  x  then
n  dx
dx
lim f xn  x   f X  x 
n 
5. Numerical Examples
Example (5.1)
The differential equation with random term in it and
random initial condition
y   Kx, y  x0   D, x   x0 , xn  ,
K, D are independent Poisson random variables with
joint PDF
Theorem 4.1
Let {Xn, n = 0, 1, ···}, {Yn, n = 0, 1, ···} be sequences of
2-r.v’s over the same probability space and let a and b be
deterministic real numbers.
Suppose: X  lim X n and Y  lim Yn
n 
fK ,D  K , D  
1) The exact solution,
n 
Then:
1)  aX  bY   lim  aX n  bYn 
n 
2) E  X   lim E  X n 
n 
y  D
n 
   EX 
2
y1  y0  hf  y0 , x0   D  hKx0
n 
Copyright © 2011 SciRes.
2
at n = 1
5) lim Var  X n   Var  X 
Definition 4.1 [13]. “The convergence in probability”

yn  yn 1  hf  yn 1 , xn 1  , y  x0   y0
n 
2
n

K x 2  x02
2) The numerical solution
Using the Random Euler Method:
3) E  XY   lim E  X nYn 
4) lim E X
e 4 2 k  D
, K , D  0,1, 2,
k !D!
at n = 2
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M. A. EL-TAWIL ET AL.
y2  y1  hf  y1 , x1   D  hKx0  hKx1
77
yn  D  nhKx0   n  n  1 2  Kh 2 .
 D  hKx0  hK  x0  h 
We can prove that:
1) lim yn  y
at n = 3
n 
y3  y2  hf  y2 , x2   D  hKx0  hKx1  hKx2
Proof
2
 D  hKx0  hK  x0  h   hK  x0  2h 
Since lim yn  y (if and only if) lim E yn  y  0
n 
n 
Then:
at n = 4
y4  y3  hf  y3 , x3   D  hKx0  hKx1  hKx2  hKx3
yn  y  nhKx0 
 D  hKx0  hK  x0  h   hK  x0  2h   hK  x0  3h 
yn  y
and so on…
Then the general numerical solution is
n  n  1
2
 h 2 K  

K x 2  x02
2
2

 n  n  1
x 2  x02
2

h 
 n h K x  2nhK x0
2
 2

2
yn  D  hKx0  hK  x0  h   hK  x0  2h 
 hK  x0  3h     hK  x0   n  1 h 
2
2
2
0
2

 n  n  1
x 2  x02
h2 
K 
2
 2

2
n 1
i.e., yn  D  hK   x0  ih  .
i 0
where h 
This can be written in another form:

 
 


2


xn  x0
n
  n  1

2
2
2
yn  y  K 2 x02  xn  x0   K 2 x0  xn  x0  
 xn  x0   x 2  x02 
 n




K 2   n  1
2
 xn  x0   x 2  x02 

4  n

2
2
 n 1 
2 2  n 1 
 K 2 x02 xn2  2 K 2 xn x03  K 2 x04  K 2 xn x0 
  xn  x0   K x0 
  xn  x0 
n
n











2
 K 2 xn x0 x 2  x02  K 2 x02 x 2  x02 

K 2  n 1 


4  n 
K 2  n 1 
K2 2
2
2
2
x
x
x
x
x  x02





n
0
0


2  n 
4




2
 xn  x0 
4
2
lim E yn  y  18 x02 x 2  36 x03 x  18 x04  18 xx0  x  x0   18 x02  x  x0 
2
2
2
n 
18 4
x  18 x3 x0  27 x 2 x02  18 xx03
4
18
18
18
 x04  9 x 4  18 x3 x0  18 xx03  9 x04  x 4  9 x 2 x02  x04
4
4
4
18
 18 x02 x 2  36 x03 x  18 x04  72 xx03  36 x 2 x02  36 x04  x 4  18 x 3 x0
4
18
 27 x 2 x02  18 xx03  x04  9 x 4  18 x 3 x0  18 xx03  9 x04
4
18 4
18 4
2 2
 x  9 x x0  x0  0
4
4




 18 xx0 x 2  x02  18 x02 x 2  x02 
i.e; lim yn  y
n 
Copyright © 2011 SciRes.
We can verify theorem (4.1) as follows
OJDM
M. A. EL-TAWIL ET AL.
78
2) lim E  yn   E  y
Proof
n 
n 1
n
i 0
j 1
yn  D  hK   x0  ih   D  hK   x0   j  1 h   D  hK  x0   x0  h    x0  2h    x0  3h     x0   n  1 h  

 n  n  1  
 D  hK  nx0  h 1  2  3     n  1    D  hK  nx0  h 

2  



 n  n  1  
xn  x0
 D  K  nhx0  h 2 
  where h 
n
2

 

2


 xn  x0    xn  x0    n  n  1  
 n  1
2

 
 D  K nx0 
 xn  x0  
 
  D  K  x0  xn  x0  
2 
2n
n
n




 



 n  1
 n  1
2 
2
 

E  yn   E  D   E  K  x0  xn  x0  
E  xn  x0  
 xn  x0     2  2  E  x0 xn   E x02 
2
n
2
n

 

 
 
Then:
1
2

lim E  yn   2  2  E  x0 x   E x02  E  x  x0  
2


1
1


 2  2  E  x0 x   E x02  E x 2  E  x0 x   E x02 
2
2


1
1

 2  2  E x 2  E x02   2  E x 2  E x02  E  y
2
2


 
n 
 
 
 
 
    
 
i.e.; lim E  yn   E  y .
n 
 
 
Since yn  D  nhKx0   n  n  1 2  Kh 2
Then we have:
3) lim E yn2  E y 2
n 
Proof
 
E yn2  E  D  Kx0  xn  x0   Kn  n  1 2   xn  x0 


2
n2 

2

2
2
 E  D  K  xn  x0  x0   2 E  D  K  xn  x0  x0   K  n  1 n   xn  x0  2  



2
 E  K  n  n  1 n   xn  x0  2  



2




2
2
 E  D 2   E  K  xn  x0  x0   2 E  DK xn x0  x02   2 E  D  K  xn  x0  x0   K  n  1 n   xn  x0  2  



2
2
 E[ K 2  n  n  1 n  E  xn  x0  2 


2
 E  D 2   E  K 2  E  xn x0   2 E  K 2  E  xn x03   E  K 2  E  x0   2 E  D  E  K  E  xn x0   2 E  D  E  K  E  x02 
2
4
2
3
 2 E  D  E  K   n  1 n  E  xn  x0  2   2  n  1 n  E  K 2  E  x0  x  x0  2 




2
 [n  n  1 n]2 E  K 2  E  xn  x0  2 


2
 E  D   E  K 2  E  xn x0   2 E  K 2  E  xn x03   E  K 2  E  x0   2 E  D  E  K  E  xn x0   2 E  D  E  K  E  x02 
2
2
4
2
1
2
3
4
 E  D  E  K   n  1 n  E  xn  x0     n  1 n  E  K 2  E  x0  x  x0     n  n  1 n  E  K 2  E  xn  x0  



 4


Copyright © 2011 SciRes.
OJDM
M. A. EL-TAWIL ET AL.
79
Then by taking the limit:
 
lim E yn2  6  6 E  xx0   12 E  xx03   6 E  x0   8E  xx0   8 E  x02 
n 
2
4
6
2
3
4
 4 E  x  x0   6 E  x0  x  x0    E  x  x0 

 4
2
4
 6  6 E  xx0   12 E  xx03   6 E  x0   8 E  xx0   8E  x02   4 E  x 2 
 8 E  xx0   4 E  x02   6 E  x3 x0   18 E  x 2 x02   18E  xx03   6 E  x04 

 
6
6
E  x 4   6 E  x3 x0   6 E  xx03   9E  x 2 x02   E  x04   E y 2
4
4
 
 
i.e. lim E yn2  E y 2 .
n 
fy  y 
4) lim var  yn   var  y
n 
Proof
 
 lim E  y   lim  E  y  
 E  y    E  y    Var  y 
2
lim Var  yn   lim  E yn2   E  yn   
n  

n 
2
n
n 
2
2
2
i.e., lim Var  yn   Var  y  .
n 
J
5) lim PDF  yn   PDF  y 
n 
Proof
Since y  D 

K x 2  x02

2
Let us define Z = D. Then the inverse transformation
is:
D Y 
K

K x 2  x02
2 y  Z 
2
 , D = Z then we have D = Z and
2
0
k
y
J
D
y
K
yn
K
Z n
D
yn
D
Z n
1
2

nhx
n


0
  n  1 2  h
0

2
2
1
nhx0   n  n  1 2  h 2
1
1
nhx0   n  n  1 2  h 2
k
2
z
2
 x  x02
D
0
z
2
x  x02 
2
1


 yn  z n 
f yn , Z n  yn , zn   f K , D 
, zn  J
2


 nhx0   n  n  1 2  h

2
x  x02
2
4
Z
2Y  Z 
 2 y  Z  
e 4 2 X  X 0
f y , z  y, z   f K , D  2
,
Z
J


2
 2 y  Z 
 x  x0

Z ! 2
!
2 
 x  x0 
Since D  0 then Z  0 hence
K x 2  x02
yZ
2

Copyright © 2011 SciRes.
2
Zn 
2
f y n  yn  
where h 
Yn  Z n 
nhx0   n  n 1 2  h 2
e 2


 yn  z n 
!
zn ! 
2 
 nhx0   n  n  1 2  h 

Then:

2Y  Z 
Then:
x x
2
4
For a numerical solution:
since yn  D  nhKx0   n  n  1 2  Kh 2
 yn  z n 
Let zn  D then K 
nhx0   n  n  1 2  h 2
n
n 
Z
e 2 x  x0
  2  y  Z    PDF  y  .
Z 0
Z ! 2
!
2 
 x  x0 
y
4
Zn 
Yn  zn 
nhx0   n  n 1 2  h 2
e 2


Zn 0
 yn  z n 
zn ! 
!
2 
 nhx0   n  n  1 2  h 
yn

xn  x0
n
OJDM
M. A. EL-TAWIL ET AL.
80
zn 
4
e 2
yn
f y n  yn   
Yn  zn 
 xn  x0  x0 [  n 1 2 n ] xn  x0 2
yn  D 
 yn  z n 
]!
2
 xn  x0  x0   n  1 2n   xn  x0 
 PDF  yn 
zn
zn ![
1
yn  D  hK  x0   x0  h    x0  2h 
2
,
1

    x0   n  1 h    x0  nh  
2

Then by taking the limit we have
Z
2 Y  z 
2
2
e 4 2 x  x0
lim f yn  yn   
 fy  y
n 
 2 y  z
Z 0
!
z ! 2
2 
 x  x0 
y
hK
 x0  2 x1  2 x2  2 x3  .......  2 xn 1  xn  ,
2
1
yn  D  hK  x0   x0  h    x0  2h 
2
1

    x0   n  1 h    x0  nh  
2

i.e.; lim PDF  yn   PDF  y 
n 
n 1
1
yn  D  hK   x0  ih   nh 2 K .
2
i 0
B. Using the Random Runge-Kutta method:
yn 1  yn 
h
 f  yn , xn   f  yn  f  yn , xn  , xn 1   ,

2
This can be written in another form:
at n = 0
yn  D  nhKx0 
h
y1  y0   f  y0 , x0   f  y0  f  y0 , x0  , x1  
2
h
hK
 D   Kx0  Kx1   D 
 x0  x1 .
2
2
n2 h2 K
.
2
We can prove that:
1) lim yn  y
n 
At n = 1
Proof
2
Since lim yn  y (if and only if) lim E yn  y  0
h
y2  y1   f  y1 , x1   f  y1  f  y1 , x1  , x2  
2
h
h
 D   Kx0  Kx1    Kx1  Kx2 
2
2
hK
 D
 x0  2 x1  x2 .
2
n 
n 



 
K x 2  x02
n2 2
yn  y  nhKx0   h K  
2
2
yn  y
2
 n2
x 2  x02
 n 2 h 2 K 2 x02  2nhK 2 x0  h 2 
2
2

At n = 2
h
 f  y2 , x2   f  y2  f  y2 , x2  , x3  
2
hK
h
 D
 x0  2 x1  x2    Kx2  Kx3 
2
2
hK
 D
 x0  2 x1  2 x2  x3 .
2
y3  y2 

 2
x 2  x02
2 n
2

K
h 
2
2

where h 
Then the general solution is:
 


2


xn  x0
n


2
2
2
yn  y  K 2 x02  xn  x0   K 2 x0  xn  x0   xn  x0   x 2  x02 


2
2
K 
2

 xn  x0   x 2  x02 

4


 K 2 x02 xn2  2 K 2 xn x03  K 2 x04  K 2 xn x0  xn  x0   K 2 x02  xn  x0 
2




 K 2 xn x0 x 2  x02  K 2 x02 x 2  x02 

Copyright © 2011 SciRes.
K2
4
 xn  x0 
4
K2
K2 2
2
x  x02
 xn  x0  x 2  x02 
2
4



2

2
OJDM
M. A. EL-TAWIL ET AL.
81
2
lim E yn  y
n 



 18 x02 x 2  36 x03 x  18 x04  18 xx0  x  x0   18 x02  x  x0   18 xx0 x 2  x02  18 x02 x 2  x02
2
2

18 4
18
18
18
x  18 x 3 x0  27 x 2 x02  18 xx03  x04  9 x 4  18 x3 x0  18 xx03  9 x04  x 4  9 x 2 x02  x04
4
4
4
4
18
 18 x02 x 2  36 x03 x  18 x04  72 xx03  36 x 2 x02  36 x04  x 4  18 x3 x0  27 x 2 x02  18 xx03
4
18 4
18 4
18
4
3
3
4
 x0  9 x  18 x x0  18 xx0  9 x0  x  9 x 2 x02  x04
4
4
4

2) lim E  yn   E  y
i.e.; lim yn  y .
n 
n 
Verification of Theorem (4.1):
Proof
n 1
yn  D  hK   x0  ih  
i 0
n
nh 2 K
nh 2 K
 D  hK   x0   j  1 h  
2
2
j 1
nh 2 K
2
 D  hK  x0   x0  h    x0  2h    x0  3h     x0   n  1 h   
 D  hK  nx0  h 1  2  3     n  1   
nh 2 K
2

 n  n  1   nh 2 K
 D  hK  nx0  h 
 
2  
2



 n  n  1   nh 2 K
x
 D  K  nhx0  h 2 
where h  n
 
2  
2


2

 x  x    x  x    n  n  1    xn
 D  K  nx0 n 0   n 0  
 
n
n
2 






 x0
n
 x0  K
2
2n

x  x  K
 n  1
2
 D  K  x0  xn  x0  
 xn  x0    n 0
n
2
2n


2
2
 xn  x0  K 
 n  1
2
 
E  yn   E  D   E  K  x0  xn  x0  
 xn  x0   

2n
2n

 

 2 E  xn  x0 
 n  1
2

E  xn  x0   
 2  2  E  x0 xn   E x02 
2n
2n


 

 
2

lim E  yn   2  2 E  x0 x   E x02  E  x  x0  2
n 
2
1
1


 2  2  E  x0 x   E x02  E x 2  E  x0 x   E x02 
2
2


 
 
 
 
 
 2  E x 2  E x02  E  y
i.e.; lim E  yn   E  y .
n 
 
 
3) lim E yn2  E y 2
n 
Copyright © 2011 SciRes.
Proof
Since yn  D  nhKx0 
n2 h2 K
then:
2
OJDM
M. A. EL-TAWIL ET AL.
82
 
E y
2
n
2

x  x  K 

 E  D  Kx0  xn  x0   n 0
2


2

 K  xn  x0 2  
 K  xn  x0 2 
  E 

 E  D  K  xn  x0  x0   2 E   D  K  xn  x0  x0  
2
2

 



2
2


 E  D 2   E  K  xn  x0  x0   2 E  DK xn x0  x02 
2
 K  xn  x0 2 
 K  xn  x0 2  


  E 
 2 E   D  K  xn  x0  x0  
2
2



 

2
 E  D 2   E  K 2  E  xn x0   2 E  K 2  E  xn x03   E  K 2  E  x0 
2
4
2
 2 E  D  E  K  E  xn x0   2 E  D  E  K  E  x02   2 E  D  E  K  E  xn  x0  2 


3
2
 2 E  K 2  E  x0  x  x0  2   E  K 2  E  xn  x0  2 




2
 E  D   E  K 2  E  xn x0   2 E  K 2  E  xn x03   E  K 2  E  x0 
2
2
4
2
 2 E  D  E  K  E  xn x0   2 E  D  E  K  E  x02   E  D  E  K  E  xn  x0  


1
3
4
 E  K 2  E  x0  xn  x0    E  K 2  E  xn  x0  

 4


 
lim E yn2  6  6 E  xx0   12 E  xx03   6 E  x0   8 E  xx0   8 E  x02 
n 
2
4
6
2
3
4
 4 E  x  x0    6 E  x0  x  x0    E  x  x0  



 4 

 6  6 E  xx0   12 E  xx03   6 E  x0   8 E  xx0   8E  x02   4 E  x 2 
2
4
 8E  xx0   4 E  x02   6 E  x3 x0   18 E  x 2 x02   18 E  xx03   6 E  x04 

 
6
6
E  x 4   6 E  x3 x0   6 E  xx03   9 E  x 2 x02   E  x04   E y 2
4
4
 
 
is:
i.e.; lim E yn2  E y 2 .
n 
4) lim Var  yn   Var  y 
D  y
n 
Proof
 
 lim E  y   lim  E  y  
 E  y    E  y    Var  y 
2
lim Var  yn   lim  E yn2   E  yn   
n 
n  

2
n
n 
2
n
n 
2
2
i.e. lim Var  yn   Var  y 
n 
5) lim PDF  yn   PDF  y 
n 
Since y  D 

K x 2  x02

2
Let us define Z = D. Then the inverse transformation
Copyright © 2011 SciRes.
K

K x 2  x02
2 y  z
2

D = z then we have D = z and
x 2  x02
K
y
J
D
y
K
2
z
2
 x  x02
D
0
z
2
x  x02 
2
1
2
x  x02
2
Z
2Y  Z 
 2 y  z 
e 4 2 X  X 0
f y , z  y, z   f K , D  2
,
z
J


2
 2 y  z
 x  x0

z ! 2
!
2 
 x  x0 
2
2
Since D  0 then z  0 this implies
OJDM
M. A. EL-TAWIL ET AL.

K x 2  x02
y z

i.e.; lim PDF  yn   PDF  y  .
n 
2
Z
Example (5.2)
Solve the problem
2Y  Z 
2
2
e 4 2 X  X 0
 PDF  y  .
 2 y  z 
z 0
z ! 2
!
2 
 x  x0 
y
fy  y  
dy
 y 2  Ky, y  0   K , t   0, tn  , K  exp 1
dt
The exact solution
Numerically,
n2 h2 K
since yn  D  nhKx0 
2
 yn  z n 
Let zn  D then K 
n2 h2
nhx0 
2
k
k
1
1
2 2
yn zn
n h
n2 h2
J
 nhx0 
nhx0 
D D
2
2
0
1
yn zn
1

nhx0 
f yn , zn
AT n = 1






y1  y0  hf  y0 , t0   K  h y02  Ky0  K .
At n = 2
y2  y1  hf  y1 , t1   K  h y12  ky1  K .
At n = 3
2
n 
e 4 2
 
zn  0
  yn  z n 
zn ! 
2 2
 nhx  n h
0

2


!


n 
E  yn   E  K   K  E  y 
4
Zn 
e 2
where h 
Yn  Z n 
 xn  x0  x0   xn  x0 2
xn  x0
n
 
Z
4) lim Var  yn   Var  y 
 
 lim E  y   lim  E  y  
 E  y    E  y    Var  y 
2
lim Var  yn   lim  E yn2   E  yn   
n 
n  

2
2Y  Z 
X 2  X 02
e 2
 fy  y
lim f yn  yn   
n 
 2 y  z
z 0
z ! 2
!
2 
 x  x0 
Copyright © 2011 SciRes.
 
2
n 


 yn  z n 
zn ! 
!
2
  xn  x0  x0   xn  x0  2 
 PDF  yn 
4
 
Since yn  K  E yn2  E  K   K 2  E y 2
2
n
n 
2
zn
y
 
3) lim E yn2  E y 2
n 


!


n 
Verification of Theorem (4.1)
It is clear that:
2) lim E  yn   E  y
Yn  Z n 
n2 h2
nhx0 
2
2
Since lim E yn  y  lim E K  K  0
n 2 h2
2
e 2

  yn  z n 
zn ! 
2 2
 nhx  n h
0

2
yn
f y n  yn   
yn  yn 1  hf  yn 1 , tn 1  , y  0   K .
n 
Yn  Z n 
nhx0 
4
Zn 
yn
The numerical solution by the Euler method:
And so on….
Then the general numerical solution: yn  K .
It is clear that:
1) lim yn  y


  yn  z n 

, zn  J
 yn , z n   f k , D 
2 2
 nhx  n h



0
2


Zn 
f y n  yn  
yK
y3  y2  hf  y2 , t2   K  h y22  ky2  K .
n2 h2
2

83
2
n
n 
2
5) lim PDF  yn   PDF  y 
n 
y  K Then J  1 which implies
PDF  y   J PDF  k   e  y
yn  K Then J  1 which implies
PDF  yn   J PDF  K   e  yn k
OJDM
M. A. EL-TAWIL ET AL.
84
Then: lim PDF  yn   PDF  y 
Euler Method for Solving Differential Equations with
Uncertainties,” Progress in Industrial Mathematics at
ECMI, Madrid, 2006.
n 
6. Conclusions
The initially valued first order random differential equations can be solved numerically using the random Euler
and random Runge-Kutta methods in mean square sense.
The existence and uniqueness of the solution have been
proved. The convergence of the presented numerical
techniques has been proven in mean square sense. The
results of the paper have been illustrated through some
examples.
[5]
H. Lamba, J. C. Mattingly and A. Stuart, “An adaptive
Euler-Maruyama Scheme for SDEs, Convergence and
Stability,” IMA Journal of Numerical Analysis, Vol. 27,
No. 3, 2007, pp. 479-506. doi:10.1093/imanum/drl032
[6]
E. Platen, “An Introduction to Numerical Methods for
Stochastic Differential Equations,” Acta Numerica, Vol. 8,
1999, pp. 197-246. doi:10.1017/S0962492900002920
[7]
D. J. Higham, “An Algorithmic Introduction to Numerical Simulation of SDE,” SIAM Review, Vol. 43, No. 3,
2001, pp. 525-546. doi:10.1137/S0036144500378302
[8]
D. Talay, and L. Tubaro, “Expansion of The Global Error
for Numerical Schemes Solving Stochastic Differential
Equation,” Stochastic Analysis and Applications, Vol. 38,
No. 4, 1990, pp. 483-509.
doi:10.1080/07362999008809220
[9]
P. M. Burrage, “Numerical Methods for SDE,” Ph.D.
Thesis, The University of Queensland, 1999.
7. References
[1]
[2]
[3]
[4]
K. Burrage, and P.M. Burrage, “High Strong Order Explicit Runge-Kutta Methods for Stochastic Ordinary Differential Equations,” Applied Numerical Mathematics,
Vol. 22, No. 1-3, 1996, pp. 81-101.
doi:10.1016/S0168-9274(96)00027-X
K. Burrage, and P. M. Burrage, “General Order Conditions for Stochastic Runge-Kutta Methods for Both
Commuting and Non-Commuting Stochastic Ordinary
Equations,” Applied Numerical Mathematics, Vol. 28, No.
2-4, 1998, pp. 161-177.
doi:10.1016/S0168-9274(98)00042-7
J. C. Cortes, L. Jodar and L. Villafuerte, “Numerical
Soluion of Random Differential Equations, a Mean
Square Approach,” Mathematical and Computer Modelling, Vol. 45, No.7, 2007, pp. 757-765.
doi:10.1016/j.mcm.2006.07.017
[10] P. E. Kloeden, E. Platen and H. Schurz, “Numerical Solution of SDE Through Computer Experiments,” Second
Edition, Springer, 1997.
[11] M. A. El-Tawil, “The Approximate Solutions of Some
Stochastic Differential Equations Using Transformation,”
Journal of Applied Mathematics and Computing, Vol.
164, No. 1, 2005, pp. 167-178.
doi:10.1016/j.amc.2004.04.062
[12] P. E. Kloeden, and E. Platen, “Numeical Solution of Stochastic Differential Equations,” Springer, Berlin, 1999.
[13] T. T. Soong, “Random Differential Equations in Science
and Engineering,” Academic Press, New York, 1973.
J. C. Cortes, L. Jodar, and L.Villafuerte, “A Random
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