Mean square Convergent Finite Difference Scheme For
Random Parabolic Differential Equations
Magdy A. El-Tawil*
Cairo University, Faculty of Engineering, Engineering
Mathematics Department, Giza, Egypt
*: corresponding author: magdyeltawil@yahoo.com
Mohammed. A. Sohaly
Mansoura University, Faculty of Science, Math.Dept,
Mansoura, Egypt. m_stat2000@yahoo.com
Abstract
In this paper, the random finite difference method is used in solving
random parabolic differential equations. 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 Partial Differential Equations (RPDEs), Mean Square Sense
(m.s), Second Order Random Variable, Random Finite Difference
Method.
1. Introduction
Random partial differential equations (RPDE) are defined as partial differential equations
involving random inputs. Various numerical methods and approximation schemes for RPDEs
have also been developed, analyzed, and tested (see [1-6]).
This paper is interested in studying the following random parabolic differential problem
of the form:
ut ( x, t )  u xx ( x, t ) , t  [0, T ] , x  (0, X )
u ( x,0)  u 0 ( x)
, x  [0, X ]
(1.1)
u (0, t )  u ( X , t )  0
Randomness may exist in the initial condition or/and in  . The random finite difference
method is used to obtain an approximate solution for problem (1.1).
This paper is organized as follows. In Section 2, some important preliminaries are
discussed. In Section 3, the Consistency of (RFDS), Stability of (RFDS) and the
Convergence of (RFDS) are discussed. Section 4 presents some results. Section 5
1
presents the solution of some numerical examples of random parabolic partial differential
equations using random finite difference method. The general conclusions are presented
in the end section.
2. Preliminaries
2.1. Mean Square Calculus
Definition1. [7]. Let us consider the properties of a class of real r.v.'s
X 11 , X 12 ,...., X 21 , X 22 ,... X nk ,.... whose second moments, E{ X 112 }, E{ X 122 },...., E{ X nk2 },...
are finite. In this case, they are called "second order random variables", (2.r.v's).
Definition2 [7]. 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 , t 2 , …… t n , the r.v's { X (t1 ), X (t 2 ),...., X (t n )}
are elements of L2 -space. A second order s.p. { X (t ), t  T } is characterized by
X (t )  E{ X 2 (t )}  
2
, t T
2.1.1 The convergence in mean square [7]
A sequence of r.v's { X nk , n, k  o } converges in mean square (m.s) to a random variable
X if:
l.i.m X nk  X =0
n , k  
m. s
X
i.e. X nk 
or L.i.m X nk = X
n , k  
where l.i.m is the limit in mean square sense.
3. Random Difference Scheme (RFDS)
In this work, the finite difference method is adapted to the random case in order to
approximate random parabolic differential equations of the following form:
ut ( x, t )  u xx ( x, t ) , t  [0, T ] , x  (0, X ) (3.1)
u ( x,0)  u 0 ( x)
, x  [0, X ]
(3.2)
u (0, t )  u ( X , t )  0
(3.3)
Consider a uniform mesh with step size x and t on x-axis and t-axis respectively.
u kn approximates u ( x, t ) at point ( kx, nt ), hence u k0  u 0 (kx) . On this mesh, we
have:
u kn 1  u kn
u ( x, t  t )  u ( x, t )
u t ( x, t ) 
u t (kx, nt ) 
, then
t
t
Similarly:
2
u ( x  x, t )  2u ( x, t )  u ( x  x, t )
u xx ( x, t ) 
, then
x 2
Hence (3.1) can be represented as follows:
u kn1  2u kn  u kn1
u xx (kx, nt ) 
x 2
u kn 1  u kn  r (u kn1  2u kn  u kn1 )
(3.4)
u  u 0 ( x)
(3.5)
u u 0
(3.6)
0
k
n
0
n
X
t
.
x 2
The tbove scheme is a random version of (3.4-3.6). For a RPDE, say Lv  G ,
where L is a differential operator and G  L2 ( R) in the other hand, we represent finite
where r  
difference scheme at the point (kx, nt ) by Lnk u kn  Gkn .
The tbove random difference scheme (3.4) can be written in the followig form:
u kn 1  (1  2r )u kn  r (u kn1  u kn1 )
3.1. Consistency Of (RFDS)
Definition 3[3]. A random difference scheme Lnk u kn  Gkn approximating RPDE Lv  G
is consistent in mean square at time t  (n  1)t , if for any continuously differentiable
function   ( x, t ) , we have in mean square:
2
E ( L  G ) nk  ( Lnk (kx, nt )  G kn )  0
As x  0 , t  0 and (kx, nt )  ( x, t )
Theorem (3.1)
The random difference scheme (3.4-3.6) is consistent in mean square sense.
Proof:
Assume that  ( x, t ) is a smooth function then:
( n 1) t
L( )
n
k
  (kx, (n  1)t )   (kx, nt )  

xx
(kx, s)ds ,
nt
and:
Lnk    (kx, (n  1)t )   (kx, nt )  r (((k  1)x, nt )  2(kx, nt )  ((k  1)x, nt ))
Then we have:
( n 1) t
2
t
n
n
E ( L ) k  Lk   E      xx (kx, s )ds  
( (( k  1)x, nt )  2 (kx, nt ) 
x 2
nt
 (( k  1)x, nt ))
 E   [
2
( n 1) t

x
(kx, s)ds 
nt
 E  [
( n 1) t

nt
x
(kx, s)ds 
t
((( k  1)x, nt )  2(kx, nt )  (( k  1)x, nt ))]
x 2
t
((( k  1)x, nt )  2(kx, nt )  (( k  1)x, nt ))]
x 2
3
2
2
( n 1) t
 [ E[  [
2
  x (kx, s)ds]  2 [(
2
( n 1) t
  x (kx, s)ds)(
2
nt
 (( k  1)x, nt )))]   2 [
(kx, nt )  ( x, t )
If
nt
k 2 t
( (( k  1)x, nt )  2 (kx, nt ) 
k 2 x 2
k t
2
(

(
k

1
)

x
,
n

t
)

2

(
k

x
,
n

t
)


((
k

1
)

x
,
n

t
))]

k 2 x 2

2
and
2
x, t  0 then E ( L ) nk  ( Lnk  (kx, nt ))  0
Hence the random difference scheme (3.4-3.6) is consistent in mean square sense.
3.2. Stability Of (RFDS)
Definition 4[3]. A random difference scheme is stable in mean square if there exists
some positive constants  ,  and constants k, b such that:
2
2
E u kn 1  kebt E u 0
For all 0  t  (n  1)t 0  x   and 0  t  
Theorem (3.2)
The random difference scheme (3.4-3.6) is stable in mean square sense.
Proof
n 1
n
n
n
Since u k  (1  2r )u k  r (u k 1  u k 1 ) then:
2
E u kn1  E (1  2r )u kn  r (u kn1  u kn1 )
2
 E[(1  2r ) 2 (u kn ) 2  2r (1  2r )(u kn )(u kn1  u kn1 )  r 2 (u kn1  u kn1 ) 2 ]
 E (u kn ) 2  4rE (u kn ) 2  4r 2 E (u kn ) 2  2rE (u kn u kn1 )  2rE (u kn u kn1 )  4r 2 E (u kn u kn1 ) 
4r 2 E (u kn u kn1 )  r 2 E (u kn1 ) 2  2r 2 E (u kn1u kn1 )  r 2 E (u kn1 ) 2
 sup[ E (u kn ) 2  4rE (u kn ) 2  4r 2 E (u kn ) 2  2rE (u kn u kn1 )  2rE (u kn u kn1 )  4r 2 E (u kn u kn1 ) 
k
4r E (u kn u kn1 )  r 2 E (u kn1 ) 2  2r 2 E (u kn1u kn1 )  r 2 E (u kn1 ) 2 ]
2
 sup E (u kn ) 2  4r sup E (u kn ) 2  4r 2 sup E (u kn ) 2  2r sup E (u kn ) 2  2r sup E (u kn ) 2  4r 2 sup E (u kn ) 2 
k
k
k
k
k
k
4r sup E (u )  r sup E (u )  2r sup E (u )  r sup E (u )
2
n 2
k
2
k
n 2
k
2
k
n 2
k
2
k
n 2
k
k
 sup E (u )
n 2
k
k
Hence:
2
sup E u kn1  sup E u kn
k
k
2
 sup E u kn
2
2
 ........................  sup E u k0 ,
k
k
then:
2
E u n 1  E u 0
2
where k=1 and b=0. Then the random difference scheme (3.4-3.6) is stable in mean
square sense.
3.3. Convergence of (RFDS)
Definition 5[3]. A random difference scheme Lnk u kn  Gkn approximating RPDE Lv  G
is convergent in mean square at time t  (n  1)t if
4
2
E u kn  u  0 ,
as x  0 , t  0 and (kx, nt )  ( x, t ) .
(3.3.1)A Stochastic Version of Lax-Richtmyer Theorem [3]
A random difference scheme Lnk u kn  Gkn approximating SPDE Lv  G is convergent in
mean square at time t  (n  1)t , if it is consistent and stable.
Theorem (3.3.2)
The random difference scheme (3.4-3.6) is convergent in mean square sense.
Proof:
2
E u kn  u  E ( Lnk ) 1 ( Lnk u kn  Lnk u )
2
Since the scheme is consistent then we have
m. s
Lnk u kn 
Lnk u .
2
Then we obtain E ( Lnk u kn  Lnk u  0
as x  0 , t  0 and (kx, nt )  ( x, t )
and since the scheme is stable then ( Lnk ) 1 is bounded.
2
Hence E u kn  u  0 As x  0 , t  0 .
Then the random difference scheme (3.4-3.6) is convergent in mean square sense.
4. Some Results
Theorem (4.1):
Let { X nk , n, k  o }, { Ynk , n, k  o } be sequences of 2-r.v's over the same probability
space and suppose that:
L.i.m X nk = X ,
L.i.m Ynk = Y
n , k  
n , k  
Then:
(i) lim E{ X nk }  E ( X )
n  
k  
(ii) lim E{ X nk2 }  E ( X 2 )
n  
k  
(iii) lim Var{ X nk }  Var ( X )
n  
k  
(v) lim PDF ( X nk )  PDF ( X )
n  
k  
Proof:
1
1
(i) From Schwarz inequality: E XY  ( E ( X 2 )) 2 E ( E (Y 2 )) 2 we have:
1
2
E{ X nk .1}  1.( E ( X ))  X nk
2
nk
Then
E{X nk }  E X nk  X nk  
In (4.1.1) put X nk  X instead of X nk then we have:
5
(4.1.1)
E{X nk  X }  E{X nk }  E{X }  E X nk  X  X nk  X
E{X nk }  E{X }  0
As n, k   then
lim E{ X nk }  E ( X )
i.e.,
n  
k  
(ii) From successive applications of the triangle and Schwarz inequalities:
E ( XY )  E ( X nk Ynk )  E ( XY )  E ( X nk Ynk )  E ( XYnk )  E ( XY )  E ( X nk Y )
 E (Y  Ynk ) X  E ( X  X nk )Y  E (( X  X nk )(Y  Ynk ))
 E (Y  Ynk ) X  E ( X  X nk )Y  E ( X  X nk )(Y  Ynk )
 X Ynk  Y  Y X nk  X  X nk  X Ynk  Y
  .
But each of the terms on the right-hand side tends to zero by hypothesis as n,k 
E ( XY ) 
Then
as X nk  Ynk . Then we obtain
(iii)Since
lim
n , k  
E{ X nk Ynk }
E{ X nk2 }  E{ X 2 } .
lim E{ X nk }  E ( X ) and
lim
n , k  
n  
k  
E{ X nk2 }  E{ X 2 }
Var{ X nk }  E{ X nk2 }  ( E{ X nk }) 2 ,
and
then
lim
n , k  
lim Var{ X nk }  lim ( E{ X nk2 }  ( E{ X nk }) 2 )
n , k 
= lim
n , k  
n , k 
E{ X nk2 } 
lim ( E{ X nk}) 2  E{ X 2 }  {E ( X )}2  Var ( X ) .
n , k  
lim Var{ X n }  Var{ X } .
Then we obtain
n , k  
(v) Definition 6.[7]. "The convergence in probability"
A sequence of r.v's { X nk } converges in probability to a random variable X
  if
as n,k 
lim
n , k  
p{ X nk  X   } =0  >0
Definition 7.[7]. "The convergence in distribution"
A sequence of r.v's { X nk } converge in distribution to a random variable X
  if
as n , k 
lim
n , k  
Fx
( x)  Fx ( x) .
nk
Lemma (4.1.1) [7]
The convergence in m.s implies convergence in probability .
Lemma (4.1.2) [7]
6
The convergence in probability implies convergence in distribution.
Theorem 4.2
m. s
m. s
X then PDF of { X nk } 
If X nk 
PDF of { X }
i.e.
lim
n,k 
 
fx
( x) 
f
nk
X
(x) .
The proof
Since we have shown that If
d
X nk 

X
m. s
X nk 
X then
m. s
X
i.e. if X nk 
then
lim
n , k  
d
d
FX nk ( x) 
FX ( x )
n , k  dx
dx
then lim
Fx
hence
( x) 
nk
lim
n,k 
 
F
fx
X
(x)
( x) 
f
nk
X
(x)
5. Numerical Examples:
Example (1):
Solve the random parabolic partial differential equation:
ut ( x, t )  u xx ( x, t ) , t  [0, T ] , x  (0, X )
u ( x,0)   sin x
 ~ exp(1)
, x  [0, X ]
u (0, t )  u ( X , t )  0
The exact solution:
(5.1)
(5.2)
(5.3)
u( x, t )  e  t sin x
2
The numerical solution:
For the difference method, consider a uniform mesh with step size x and t on x-axis
X
T
and t-axis respectively, where x 
, t 
and M , N  0 .
M
N
u kn approximates u ( x, t ) at point ( kx, nt ), u k0  u 0 (kx) . On this mesh, the difference
scheme for this problem is:
u kn 1  (1  2r )u kn  r (u kn1  u kn1 )
(5.4)
u k0   sin x
(5.5)
u u 0
(5.6)
n
0
n
X
t
, xk  kx , t n  nt .
x 2
First from the initial condition we have:
u(0,0)  u 00  0
where r 
u(1,0)  u10   sin( x )
u(2,0)  u 20   sin( 2x )
then:
u(k ,0)  u k 0   sin( kx ) .
From (5.4):
u(1,1)  u11  (1  2r )u10  r (u 20  u 00 )
7
(5.7)
 (1  2r )u10  ru 20
  (1  2r )[sin( x )]  r [sin( 2x )]
u(2,1)  u 21  (1  2r )u 20  r (u30  u10 )
  (1  2r )[sin( 2x )]  r[sin( 3x )  sin( x )]
Finally:
u (k ,1)  u k1   (1  2r )[sin( kx )]  r [sin( k  1)x  sin( k  1)x ]
  (1  2r )[sin( kx )]  2r [sin( kx ) cos( x )]
  sin( kx )[1  2r  2r cos( x )]   sin( kx )[1  2r (1  cos( x )]
x
x
  sin( kx )[1  2r (2 sin 2 (
)]   sin( kx )[1  4r sin 2 (
)]
2
2
  sin( kx )[1  4r (
e
ix
2
  sin( kx )[1  r (e ix
 ix
e 2 2
) ]   sin( kx )[1  r (e
2i
 2  e ix )]
  sin( kx )[1  r (2  1  ix 
ix
2
e
 ix
2
)2 ]
(x ) 2 i (x ) 3 (x ) 4


 .........  1  ix 
2!
3!
4!
(x ) 2 i (x ) 3 (x ) 4


 .........)]
2!
3!
4!
2(x ) 2 2(x ) 4 2(x ) 2
  sin( kx )[1  r (


 ........)]
2!
4!
6!

(x ) 2i
  sin( kx )[1  2r  (1) i
)]
(2i)!
i 1
Then:
2i

i (x )
u k1   sin( kx )[1  2r  (1)
)]
(2i)!
i 1
u (1,2)  u12  (1  2r )u11  r (u 21  u 01 )

 (1  2r )  sin( x )[1  2r  (1) i
i 1
(5.8)

(x ) 2i
(x ) 2i
]  r sin 2x [1  2r  (1) i
]
(2i)!
(2i)!
i 1
u (2,2)  u 22  (1  2r )u 21  r (u31  u11 )

 (1  2r )  sin( 2x )[1  2r  (1) i
i 1

(x ) 2i
(x ) 2i
]  r sin 3x [1  2r  (1) i
]
(2i)!
(2i)!
i 1
(x ) 2i
]
(2i)!
i 1
u (3,2)  u32  (1  2r )u31  r (u 41  u 21 )

 r sin x [1  2r  (1) i

(x ) 2i
(x ) 2i
]  r sin 4x [1  2r  (1) i
]
(2i)!
(2i)!
i 1
i 1

(x ) 2i
 r sin 2x [1  2r  (1) i
]
(2i)!
i 1

 (1  2r )  sin( 3x )[1  2r  (1) i
8
Finally:

(x ) 2i
(x ) 2i
]  r [sin( k  1)x ][1  2r  (1) i
]
(2i)!
(2i)!
i 1
i 1

(x ) 2i
 r [sin( k  1)x ][1  2r  (1) i
]
(2i)!
i 1

u(k ,2)  u k 2  (1  2r ) sin( kx )[1  2r  (1) i

u (k ,2)  u k 2  (1  2r )  sin( kx )[1  2r  (1) i
i 1

(x ) 2i
(x ) 2i
]  r [1  2r  (1) i
]
(2i )!
(2i)!
i 1
[sin( k  1)x  sin( k  1)x ]

u (k ,2)  u k 2  (1  2r )  sin( kx )[1  2r  (1) i
i 1

(x ) 2i
(x ) 2i
]  2r [1  2r  (1) i
]
(2i)!
(2i)!
i 1
[sin( kx ) cos( x )]

  sin( kx )[1  2r  (1) i
i 1

  sin( kx )[1  2r  (1) i
i 1
(x ) 2i
][1  2r  2r cos( x )]
(2i)!
(x ) 2i 2
]
(2i)!
Then:

u k 2   sin( kx )[1  2r  (1) i
i 1
(x ) 2i 2
]
(2i)!
(5.9)
u(1,3)  u13  (1  2r )u12  r (u 22  u 02 )

(x ) 2i 2
(x ) 2i 2
]  r sin 2x [1  2r  (1) i
]
(2i)!
(2i)!
i 1
i 1
u(2,3)  u 23  (1  2r )u 22  r (u32  u12 )

 (1  2r )  sin( x )[1  2r  (1) i
2i

(x ) 2i 2
i (x )
 (1  2r ) sin( 2x )[1  2r  (1)
]  r sin 3x [1  2r  (1)
]2
(2i)!
(2i)!
i 1
i 1
2i

(x ) 2
 r sin x [1  2r  (1) i
]
(2i)!
i 1
u(3,3)  u33  (1  2r )u32  r (u 42  u 22 )

i

(x ) 2i 2
(x ) 2i 2
]  r sin 4x [1  2r  (1) i
]
(2i)!
(2i)!
i 1
i 1

(x ) 2i 2
 r sin 2x [1  2r  (1) i
]
(2i)!
i 1
Finally:


(x ) 2i 2
(x ) 2i 2
u k 3  (1  2r )  sin( kx )[1  2r  (1) i
]  r [sin( k  1)x ][1  2r  (1) i
]
(2i)!
(2i)!
i 1
i 1

(x ) 2i 2
 r [sin( k  1)x ][1  2r  (1) i
]
(2i)!
i 1

 (1  2r ) sin( 3x )[1  2r  (1) i
9

(x ) 2i 2
(x ) 2i 2
]  r [1  2r  (1) i
]
(2i)!
(2i)!
i 1

 (1  2r )  sin( kx )[1  2r  (1) i
i 1
[sin( k  1)x  sin( k  1)x ]

(x ) 2i 2
(x ) 2i 2
]  2r [1  2r  (1) i
]
(2i )!
(2i )!
i 1

 (1  2r )  sin( kx )[1  2r  (1) i
i 1
[sin( kx ) cos( x )]

  sin( kx )[1  2r  (1) i
i 1

  sin( kx )[1  2r  (1) i
i 1
(x ) 2i 2
] [1  2r  2r cos( x )]
(2i)!
(x ) 2i 3
]
(2i)!
Then:

u k 3   sin( kx )[1  2r  (1) i
i 1
(x ) 2i 3
]
(2i )!
(5.10)
(x ) 2i n
]
(2i)!
(5.11)
Finally the numerical solution for this problem is:

u kn   sin( kx )[1  2r  (1) i
i 1
We can prove that:
m. s
u
(i) u kn 
Proof
l.i.m u kn 
Since:
n  
k  
l.i.m
x , t  0
kx , nt  x ,t
u kn  u (if and only if)
lim
x , t  0
kx , nt  x ,t
2
lim E u kn  u  0 or
n  
k  
2
E u kn  u  0
2i

2
u kn  u   sin( kx )[1  2r  (1) i (x ) ] n  e  t sin x
(2i)!
i 1
2
u kn  u 

 sin( kx )[1  2r  (1) i
i 1

 [  sin( kx )[1  2r  (1) i
i 1
 2 t
(x ) 2i n
sin x
]  e
(2i)!
2

(x ) 2i n 2
(x ) 2i n
] ]  2 [  sin( kx )[1  2r  (1) i
] ]
(2i )!
(2i)!
i 1
[e  t sin x]  [e  t sin x]2
2
2

 [  2 sin 2 (kx )][1  2r  (1) i
i 1

(x ) 2i 2 n
(x ) 2i n
]  2 [  sin( kx )[1  2r  (1) i
] ]
(2i )!
(2i)!
i 1
[e  t sin x]  [ 2 e 2 t sin 2 x]2
2
2
Then:
10
2
E u kn  u 

E[[  2 sin 2 (kx )][1  2r  (1) i
i 1

(x ) 2i 2 n
(x ) 2i n
]  2 [  sin( kx )[1  2r  (1) i
] ]
(2i)!
(2i)!
i 1
[e  t sin x]  [ 2 e 2 t sin 2 x]2 ]
2
2
At time t  (n  1)t then:
2
E u kn  u 
E[[  2 sin 2 (kx )][1  2
[e 
2
( n 1) t
t
x 2

 (1)
i 1
sin x]]  E[ [ 2e2

 E[[  2 sin 2 (kx )][1  2t  (1) i
i 1
[e 
2
i
( n 1) t
(x ) 2i 2 n
t
] ]  2 E[ [  sin( kx )[1  2 2
(2i)!
x
2
( n1) t
2
(x ) 2i n
] ]
(2i )!
i
(x ) 2i n
] ]
(2i )!
i 1
sin 2 x]]
(x ) 2i 2 n
t
] ]  2 E[ [  sin( kx )[1  2 2
2
x (2i)!
x
sin x]]  E[ [ 2 e 2
i

 (1)
( n 1) t

 (1)
i 1
sin 2 x]]
By taking the limit as x  0 , t  0 and (kx, nt )  ( x, t )
Then we obtain:
2
E u kn  u  0
Then:
m. s
u kn 
u
Example (2):
Solve the random parabolic partial differential equation:
ut ( x, t )  u xx ( x, t ) , t  [0, T ] , x  (0, X )
u ( x,0)  sin x
 ~ poisson (1)
, x  [0, X ]
u (0, t )  u ( X , t )  0
The exact solution:
(5.12)
(5.13)
(5.14)
u( x, t )  e   t sin x
2
The numerical solution:
For the difference method, consider a uniform mesh with step size x and t on x-axis
X
T
and t-axis respectively, where x 
, t 
and M , N  0 .
M
N
On this mesh, the difference scheme for this problem is:
11
u kn 1  (1  2r )u kn  r (u kn1  u kn1 )
(5.15)
u  sin x
(5.16)
u 0n  u Xn  0
(5.17)
0
k
where r 
t
, xk  kx , t n  nt
x 2
First from the initial condition we have:
u(0,0)  u 00  0
u (1,0)  u10  sin( x )
u (2,0)  u 20  sin( 2x )
Then:
u (k ,0)  u k 0  sin( kx )
(5.18)
From (5.15):
u(1,1)  u11  (1  2r )u10  r (u 20  u 00 )
 (1  2r )u10  ru 20
 (1  2r )[sin( x )]  r[sin( 2x )]
u(2,1)  u 21  (1  2r )u 20  r (u30  u10 )
 (1  2r )[sin( 2x )]  r[sin( 3x )  sin( x )]
Finally:
u (k ,1)  u k1  (1  2r )[sin( kx )]  r[sin( k  1)x  sin( k  1)x ]
 (1  2r )[sin( kx )]  2r[sin( kx ) cos( x )]
 sin( kx )[1  2r  2r cos( x )]  sin( kx )[1  2r (1  cos( x )]
x
x
 sin( kx )[1  2r (2 sin 2 (
)]  sin( kx )[1  4r sin 2 (
)]
2
2
 sin( kx )[1  4r (
e
ix
2
 sin( kx )[1  r (e ix
 ix
2
e
) 2 ]  sin( kx )[1  r (e
2i
 2  e ix )]
 sin( kx )[1  r (2  1  ix 
ix
2
e
 ix
2
)2 ]
(x ) 2 i (x ) 3 (x ) 4


 .........  1  ix 
2!
3!
4!
(x ) 2 i (x ) 3 (x ) 4


 .........)]
2!
3!
4!
2(x ) 2 2(x ) 4 2(x ) 2
 sin( kx )[1  r (


 ........)]
2!
4!
6!

(x ) 2i
 sin( kx )[1  2r  (1) i
)]
(2i)!
i 1
Then:

(x ) 2i
u k1  sin( kx )[1  2r  (1) i
)]
(2i)!
i 1
12
(5.19)
u (1,2)  u12  (1  2r )u11  r (u 21  u 01 )

 (1  2r ) sin( x )[1  2r  (1) i
i 1

(x ) 2i
(x ) 2i
]  r sin 2x [1  2r  (1) i
]
(2i)!
(2i)!
i 1
u (2,2)  u 22  (1  2r )u 21  r (u31  u11 )
2i

(x ) 2i
i (x )
 (1  2r ) sin( 2x )[1  2r  (1)
]  r sin 3x [1  2r  (1)
]
(2i)!
(2i)!
i 1
i 1

(x ) 2i
 r sin x [1  2r  (1) i
]
(2i)!
i 1
u (3,2)  u32  (1  2r )u31  r (u 41  u 21 )

i
2i

(x ) 2i
i (x )
 (1  2r ) sin( 3x )[1  2r  (1)
]  r sin 4x [1  2r  (1)
]
(2i)!
(2i)!
i 1
i 1

(x ) 2i
 r sin 2x [1  2r  (1) i
]
(2i)!
i 1
Finally:


(x ) 2i
(x ) 2i
u k 2  (1  2r ) sin( kx )[1  2r  (1) i
]  r[sin( k  1)x ][1  2r  (1) i
]
(2i)!
(2i)!
i 1
i 1

(x ) 2i
 r[sin( k  1)x ][1  2r  (1) i
]
(2i)!
i 1

i

u (k ,2)  u k 2  (1  2r ) sin( kx )[1  2r  (1) i
i 1

(x ) 2i
(x ) 2i
]  r[1  2r  (1) i
]
(2i )!
(2i)!
i 1
[sin( k  1)x  sin( k  1)x ]

u (k ,2)  u k 2  (1  2r ) sin( kx )[1  2r  (1) i
i 1

(x ) 2i
(x ) 2i
]  2r[1  2r  (1) i
]
(2i )!
(2i)!
i 1
[sin( kx ) cos( x )]

 sin( kx )[1  2r  (1) i
i 1

 sin( kx )[1  2r  (1) i
i 1
(x ) 2i
][1  2r  2r cos( x )]
(2i )!
(x ) 2i 2
]
(2i )!
Then:

u k 2  sin( kx )[1  2r  (1) i
i 1
(x ) 2i 2
]
(2i )!
(5.20)
u(1,3)  u13  (1  2r )u12  r (u 22  u 02 )
2i

(x ) 2i 2
i (x )
 (1  2r ) sin( x )[1  2r  (1)
]  r sin 2x [1  2r  (1)
]2
(2i)!
(2i)!
i 1
i 1
u(2,3)  u 23  (1  2r )u 22  r (u32  u12 )

i
13

 (1  2r ) sin( 2x )[1  2r  (1) i
i 1

(x ) 2i 2
(x ) 2i 2
]  r sin 3x [1  2r  (1) i
]
(2i)!
(2i)!
i 1
(x ) 2i 2
 r sin x [1  2r  (1)
]
(2i)!
i 1
u(3,3)  u33  (1  2r )u32  r (u 42  u 22 )

i

(x ) 2i 2
(x ) 2i 2
]  r sin 4x [1  2r  (1) i
]
(2i)!
(2i)!
i 1
i 1
2i

i (x )
 r sin 2x [1  2r  (1)
]2
(2i)!
i 1

 (1  2r ) sin( 3x )[1  2r  (1) i
Finally:
u(k ,3)  u k 3
2i

(x ) 2i 2
i (x )
 (1  2r ) sin( kx )[1  2r  (1)
]  r[sin( k  1)x ][1  2r  (1)
]2
(2i)!
(2i)!
i 1
i 1
2i

(x ) 2
 r[sin( k  1)x ][1  2r  (1) i
]
(2i)!
i 1

i

 (1  2r ) sin( kx )[1  2r  (1) i
i 1

(x ) 2i 2
(x ) 2i 2
]  r[1  2r  (1) i
]
(2i)!
(2i)!
i 1
[sin( k  1)x  sin( k  1)x ]

 (1  2r ) sin( kx )[1  2r  (1) i
i 1

(x ) 2i 2
(x ) 2i 2
]  2r[1  2r  (1) i
]
(2i )!
(2i )!
i 1
[sin( kx ) cos( x )]
(x ) 2i 2
] [1  2r  2r cos( x )]
(2i )!

 sin( kx )[1  2r  (1) i
i 1
(x ) 2i 3
]
(2i )!

 sin( kx )[1  2r  (1) i
i 1
Then:

u k 3  sin( kx )[1  2r  (1) i
i 1
(x ) 2i 3
]
(2i)!
(5.21)
Finally, the numerical solution for this problem is:

u kn  sin( kx )[1  2r  (1) i
i 1
(x ) 2i n
]
(2i )!
(5.22)
We can prove that:
m. s
u
(i) u kn 
Proof
Since:
l.i.m u kn 
n  
k  
l.i.m
x , t  0
kx , nt  x ,t
u kn  u (if and only if)
lim
x , t  0
kx , nt  x ,t
14
2
E u kn  u  0
2
lim E u kn  u  0 or
n  
k  
2i

2
u kn  u  sin( kx )[1  2r  (1) i (x ) ] n  e   t sin x
(2i )!
i 1

2
u kn  u 
sin( kx )[1  2r  (1) i
i 1

 [sin( kx )[1  2r  (1) i
i 1
  2t
(x ) 2i n
sin x
]  e
(2i )!
2

(x ) 2i n 2
(x ) 2i n
] ]  2 [sin( kx )[1  2r  (1) i
] ]
(2i )!
(2i )!
i 1
[e   t sin x]  [e   t sin x]2
2
2

 [sin 2 (kx )][1  2r  (1) i
i 1

(x ) 2i 2 n
(x ) 2i n
]  2 [sin( kx )[1  2r  (1) i
] ]
(2i )!
(2i )!
i 1
[e   t sin x]  [e 2  t sin 2 x]2
2
2
Then:
2
E u kn  u 

E[[sin 2 (kx )][1  2r  (1) i
i 1

(x ) 2i 2 n
(x ) 2i n
]  2 [sin( kx )[1  2r  (1) i
] ]
(2i )!
(2i )!
i 1
[e   t sin x]  [e 2  t sin 2 x]2 ]
2
2
At time t  (n  1)t then:
2
E u kn  u 
E[[sin 2 (kx )][1  2
[e  
2
( n 1) t
t
x 2

 (1)
i 1
sin x]]  E[ [e 2 

 E[[sin 2 (kx )][1  2t  (1) i
i 1
[e  
2
( n 1) t
i
2
(x ) 2i 2 n
t
] ]  2 E[ [sin( kx )[1  2 2
(2i )!
x
( n 1) t
2
(x ) 2i n
] ]
(2i)!
 (1) i
(x ) 2i n
] ]
(2i)!
i 1
sin 2 x]]
(x ) 2i 2 n
t
] ]  2 E[ [sin( kx )[1  2 2
2
x (2i )!
x
sin x]]  E[ [e 2 
i

 (1)
( n 1) t
sin 2 x]]
By taking the limit as x  0 , t  0 and (kx, nt )  ( x, t )
Then we obtain:
2
E u kn  u  0
Then:
m. s
u kn 
u
15

i 1
6. Conclusions and future works
The random parabolic differential equations can be solved numerically using the
random difference method in mean square sense. Under some conditions, the
convergence of the solution scheme to the exact one is proved.
7. References.
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Stochastic approximation of some linear stochastic partial differential equations,
Stochastics Rep., 64 (1998),pp. 117–142.
[2] Ames, W.F. Numerical Methods for Partial Differential Equations (2nded.)
Barnes and Noble Inc., New York. (1992).
[3] Andrea Barth, Stochastic partial differential equations: Approximations and
applications, Ph.D. thesis, University of Oslo, CMA, September 2009..
[4] Andrea Barth and Annika Lang, Almost sure convergence of a
Galerkin{Milstein approximation for stochastic partial differential equations
driven, submitted, 2009.
[5] Andrea Barth, A finite element method for martingale-driven stochastic
partial differential equations,Comm. Stoch. Anal. 4 (2010), no. 3
[6] Bennaton. J.F., Discrete time Galerkin approximation to the nonlinear
filtering solution, J.Math. Anal. Appl., 110 (1985), pp. 364–383.
[7] Davie. A. and Gaines. J., Convergence of numerical schemes for the
solution of the parabolic stochastic partial differential equations, Math Comp., 70
(2001), pp. 121–134.
[8] Zhang. T., Numerical Approximations of Stochastic Partial Differential Equations,
Phil .M, thesis, Hong Kong University of Science and Technology, Hong Kong,
2000.
[9] Roth, c., approximations of solution of a first order stochastic partial differential
equation, Report, institute optimierung und stochastic, Universitat Halle-wittenberg,
Halle 2001.
[10] Roth, c, Difference methods for stochastic partial differential equations, Z.Zngew.
.Math Mech,(82),821-830, 2002.
[11] T.T. Soong, Random Differential Equations in Science and Engineering, Academic
Press, New York, 1973.
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