International Journal of Trend in Scientific Research and Development (IJTSRD)
Volume 3 Issue 5, August 2019 Available Online: www.ijtsrd.com e-ISSN: 2456 – 6470
Errors in the Discretized Solution of a Differential Equation
Wai Mar Lwin1, Khaing Khaing Wai2
1Faculty
of Computing, 2Department of Information Technology Support and Maintenance
1,2University of Computer Studies, Mandalay, Myanmar
How to cite this paper: Wai Mar Lwin |
Khaing Khaing Wai "Errors in the
Discretized Solution of a Differential
Equation" Published
in
International
Journal of Trend in
Scientific Research
and Development
(ijtsrd), ISSN: 24566470, Volume-3 |
IJTSRD27937
Issue-5,
August
2019,
pp.2266-2272,
https://doi.org/10.31142/ijtsrd27937
ABSTRACT
We study the error in the derivatives of an unknown function. We construct
the discretized problem. The local truncation and global errors are discussed.
The solution of discretized problem is constructed. The analytical and
discretized solutions are compared. The two solution graphs are described by
using MATLAB software.
KEYWORDS: Differential Equations, MATLAB, Heat Equation
1. INTRODUCTION
1.1. Measuring Errors
In order to discuss the accuracy of a numerical solution, it is necessary to
choose a manner of measuring that error. It may seem obvious what is meant by
the error, but as we will see there are often many different ways to measure the
error which can sometimes gives quite different impressions as to the accuracy
of an approximate solution.
Copyright © 2019 by author(s) and
International Journal of Trend in Scientific
Research and Development Journal. This
1.1.1. Errors in a Scalar Value
is an Open Access article distributed
First we consider a problem in which the answer is a single value z  R .
under the terms of
Consider, for example, the scalar ODE
the
Creative
(1.1)
u(t )  f (u (t )),u(0)  
Commons Attribution
License
(CC
BY
4.0)
and suppose we are trying to compute the solution at some particular time T, so
(http://creativecommons.org/licenses/by z=u( T). Denote the computed solution by
/4.0)
  0 . This is slightly stronger than the previous statement,
ẑ . Then the error in this computed solution is
and
means that f ( ) decays to zero faster than g ( ) . If
(1.2)
E  zˆ  z
f ( )  o( g ( )) then f ( )  O( g ( )) though the converse may
1.1.2. Absolute Error
not be true. Saying that f ( )  o(1) simply means that the
A natural measure of this error would be the absolute value
f ( )  0 as   0 .
of E,
(1.3)
E  zˆ  z
Examples 1.1.1
1
2 3
2 3  O( 2 ) as   0 , since 2  2  1 for all   .
This is called the absolute error in the approximation.

1.1.3. Relative Error
The error defined by zˆ  z is called relative error.
z
It is also sometimes convenient to use the “little-oh”
notation f ( )  O( g ( )) as   0 . This means that f ( )  0 as
g ( )
Unique Paper ID – IJTSRD27937 |
for all   1 .
2
3
If there is some constant C such that f ( )  C for all 
g ( )
sufficiently small, or equivalently, if we can bound
f ( )  C g ( ) for all  sufficiently small.
|
2
3 5
sin( )  O( ) as   0 ,since sin( )      ...   for all
1.1.4. “Big-oh” and “little-oh” notation
In discussing the rate of convergence of a numerical method
we use the notation O( p ) , the so-called “big-oh” notation. If
f ( ) and g ( ) are two functions of  then we say that
f ( )  O( g ( )) as   0 .
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2
2  o( ) as   0 , since 2  0
3
5
  0.
sin( )     O( 2 ) .
sin( )    o( ) as   0 , since

1.1.5. Taylor Expansion
Each of the function values of u can be expanded in a Taylor
series about the point x, as e.g.,
1
1
(1.4)
u ( x   )  u ( x )  u ( x )   2 u ( x )   3 u ( x )  O ( 4 )
2
6
1
1
u ( x   )  u ( x )  u ( x )   2 u ( x )   3 u ( x )  O ( 4 )
2
6
(1.5)
1.2. Finite Difference Approximations
Our goal is to approximate solutions to differential equation,
i.e. to find a function (or some discrete approximation to this
function) which satisfies a given relationship between
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various of its derivatives on some given region of space and/
or time along with some boundary conditions along the edges
of this domain. A finite difference method proceeds by
replacing the derivatives in the differential equations by finite
difference approximations. This gives a large algebraic
system of equations to be solve in place of the differential
equation, something that is easily solved on a computer.
We first consider the more basic question of how we can
approximate the derivatives of a known function by finite
difference formulas based only on values of the function itself
at discrete points. Besides providing a basis for the later
development of finite difference methods for solving
differential equations, this allows us to investigate several
key concepts such as the order of accuracy of an
approximation in the simplest possible setting.
Let u(x) represent a function of one variable that will always
be assumed to be smooth, defined bounded function over an
interval containing a particular point of interest x,
1.2.1. First Derivatives
Suppose we want to approximate u (x) by a finite difference
approximation based only on values of u at a finite number of
points near x. One choice would be to use
  u ( x) 

1
 u ( x)  u ( x)  O( 2 )
2
(1.6)
for some small values of  . It is known as forward
difference approximation.
Another one-sided approximation would be

1.2.3. Higher Order Derivatives
Finite difference approximations to higher order derivatives
can be obtained.
1
(1.10)
   2 u ( x)  3 [u ( x  2 )  3u ( x   )  3u ( x)  u ( x   )]

1
 u ( x)  u ( x)  O( 2 )
2
 u ( x )  O ( )
The first one equation (1.10) is un-centered and first order
accurate:
1
   u ( x) 
[u ( x  2 )  2u ( x   )  2u ( x   )  u ( x  2 )] (1.11)
0


2 3
1
 u ( x)   2 u ( x)  O( 4 )
4
 u ( x )  O( 2 )
2. Comparison Of Analytical and Discretized Solution
Of Heat Equation
2.1 Solutions for the Heat Equation
2.1.1. Finite Difference Method
We will derive a finite difference approximation of the
following initial boundary value problem:
ut  u xx for x  (0,1), t  0
u ( x,0)  f ( x) for x  (0,1)
1
 u ( x)  u ( x)  O( 2 )
2
(1.7)
for two different points.It is known as backward difference
approximation.
Another possibility is to use the centered difference
approximation
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1
(n  1)
The grid points in the x-direction are given by
x j = j for j=0,1,…,n+1. Similarly, we define tm=mh for
following
approximations
(1.8)
1.2.2. Second Order Derivatives
The standard second order centered approximation is
given by
u ( x   )  2u ( x)  u ( x   )
(1.9)
 2 u ( x) 
2

x   
v mj denote an approximation of u(xj,tm). We have the
1
 [  u( x)   u ( x)]
2
1
 u ( x)   2 u   O( 4 )
6
Let n  0 be a given integer, and define the grid spacing in the
x-direction by
integers m  0 , where h  t denotes the time step. Then, we
let
u(x  )  u(x  )
0u(x) 
2
|
 u ( x )  O ( 2 )
u (0, t )  u (1, t )  0 for t  0 (2.1)
u ( x)  u ( x   )
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1 2
 u ( x)  O( 4 )
12
This second equation (1.12) is second order accurate.
u ( x   )  u ( x)
  u ( x) 
 u ( x) 
u t ( x, t ) 
u ( x , t  h )  u ( x, t )
 O ( h)
h
and
u xx ( x, t ) 
u ( x   , t )  2u ( x, t )  u ( x   , t )
2
(2.2)
 O( 2 )
(2.3)
These approximations motivate the following scheme:
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v mj1  v mj
h

v mj1  2v mj  v mj1

2
for j=1,…,n, m  0
(2.4)
Tm1  (1  h )Tm for m  0 .
Some iterations of (2.12)
By using the boundary conditions of (2.1), we have
v0m  0 and v nm1  0 , for all m  0 .
Tm1  (1  h )Tm
The scheme is initialized by
.
.
.
 (1  h ) m1 T0
v 0j  f ( x j )
 (1  h ) 2 Tm 1
, for j=1,…,n.
h
r 2
 . Then the scheme can be rewritten in a more
Let
convenient form
v mj1  rv mj1  (1  2r )v mj  rv mj1
, j=1,…,n, m  0
(2.5)
When the scheme is written in this form, we observe that the
values on the time level tm+1 are computed using only the
values on the previous time level tm and we have to solve a
tridiagonal system of linear equations.
2.1.2 Approximate Solution
The first step in our discretized problem is to derive a family
of particular solutions of the following problem:
v
m 1
j
v
m
j
h

v
m
j 1
 2v  v

m
j
2
 (1  h ) m 1
Clearly indicate that the solution is
Tm  (1  h ) m for m  0 .
v nm1  0
and
(2.6)
, for all m  0 .
(2.7)
The initial data will be taken into account later. We seek
particular solutions of the form
v  X j Tm
m
j
2
k 

for j=1,…,n, m  0

2
{Tm }m0 is a sequence of real numbers. By inserting (2.8)
into (2.6), we get
X j Tm1  X jTm X j 1Tm  2 X jTm  X j 1Tm

h
2
2
cos( )
(1  cos( ))
2
2
2 sin 2 (
2
k 
(2.8)
Here X is a vector of n components, independent of m, while

2 2
m
j 1
for j=1,…,n, m  0
4
2

2
2 sin 2 (
that
, and thus we obtain
Tm 1  Tm X j 1  2 X j  X j 1

hTm
 2X j
.

, for j=1,…,n,
Tm 1  Tm
  Tm
h
for m  0 .
X  , j  sin(jx j )
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


(2.10)

(2.11)
We first consider the equation (2.10). We define T0=1 and
consider the difference equation
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for   1,..., n
(2.14)
X k  ( X  ,1 , X  , 2 ,..., X  ,n )  R n ,   1,..., n
(2.9)
We also derive from the boundary condition (2.7) that
X 0  X n 1  0
2
, j=1,…,n.
It can be easily verified that
1
2
1
AX  , j  2 X  , j 1  2 X  , j  2 X  , j 1



The left-hand side only depends on m and the right-hand side
only depends on j. Consequently, both expressions must be
equal to a common constant, say ( ) , and we get the
following two difference equations:
2

and the corresponding eigenvectors
X jTm  0
  X j
)
have components given by
X  , j  sin(j )
Since we are looking only for nonzero solutions, we assume
X j 1  2 X j  X j 1
(2.13)
This fact is easily verified by induction on m. Next we turn
our attention to the problem (2.9) with boundary condition
(2.11). In fact this is equivalent to the eigenvalue problem.
 ,  ,...,  n are given
Hence, we obtain that the n eigenvalues 1 2
by
with the boundary conditions
v0m  0
(2.12)

1

1
2
sin( ( j  1) ) 
2

sin(j   )) 
2
sin(j ) 
2
2
1
2
sin(j ) 
sin( ( j  1) )
1
2
sin(j   )
1  sin(j ) cos( )  cos(j ) sin( )  2 sin(j ) 


 2   sin(j ) cos( )  cos(j ) sin( )

2
2
2

2
2
sin(j )(1  cos( ))
sin(j )(2 sin 2 (
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
4
2
sin 2 (

2
   ( ) 2 for   1,2,.....
) sin(j )
(2.20)
with the eigenfunctions
X  ( x)  sin(x) for   1,2,.....
  X  , j
Hence, we obtain particular solutions
vm, j  (1  h  ) m sin(x j )
m
, j
v
of the form
(2.15)
We have derived a family of particular solutions
{v }n 1 with
vm
values  , j at the grid point (xj,tm). Next, we observe that
any linear combination of particular solutions
(2.21)
 ( ) 2 t

Further, the solution of T   T is given by T (t )  e
for
  1,2,.....
(2.22)
Finally, we use the superposition principle to a solution to the
initial-boundary value problem (2.1), and we get

u ( x, t )   c T (t ) X  ( x)
 1
n
v     v
 1

, (   is scalar)
u ( x, t )   c e  ( ) t sin(x )
 1
is also a solution of (2.6) and (2.7). Finally, we determine the
coefficients {   } by using the initial condition
v 0j  f ( x j )
The unknown coefficients can be determined from the initial
condition such that

, for j=1,…,n.
Since v  X  at t=0 , we want to determine {   } such that
n
 X


1
 f (x j )
,j
f ( x )   c sin(x )
(2.23)
 1
The solution of the continuous problem is given by
, for j=1,…,n.
(2.16)

u ( x, t )   c e   t sin(x )
(2.24)
 1
Hence, it follows from
n
f ( x j )     sin(x j )
2
where   ( ) fourier coefficient
 1
1
c  2  f ( x) sin(x)dx
n
X sj , f ( x j )
2
h
    sin x j , sin sx j
 1
h
2.2. Comparison of Analytical and Discretized Solution
2.2.1. We want to compare this analytical solution with
the discretized solution given by
that
n
   2  f ( x j ) X  , j
j 1
(2.25)
0
n
v mj     (1  h ) m sin(x j )
for   1,..., n
(2.26)
 1
2.1.3. Exact Solution
To find a solution of the initial-boundary value problem (2.1),
assume that
u ( x, t )  X ( x)T (t )
(2.17)
where
4
  
  2 sin 2 


 2 
Using boundary conditions we get
and
X(0) = X(1) = 0
   2  f ( x j ) sin(x j )
(2.27)
n
j 1
If we insert the (2.17) in the equation (2.1), we have
X ( x) T(t )

X ( x ) T (t )
(2.18)
for   1,..., n
(2.28)
In order to compare the analytical and discretized solution at
a grid point (xj,tm), we define
u mj  u ( x j , t m )
, i.e,

Now we observe that the left hand side is a function of x ,
while the right hand side just depends on t. Hence, both have
to be equal to the same constant    R . This yields the
eigenvalue problem for x
X     X , X(0) = X(1) = 0
(2.19)
Nontrivial solutions only exist for special values of 
They are so-called the eigenfunctions. In this special case we
have the eigenvalues
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u mj   c e   t m sin(x j )
 1
(2.29)
Our aim is to prove
v mj  u mj
under appropriate conditions on the mesh parameters  and
h. To avoid technicalities, we consider a fixed grid point
(xj,tm) where
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parameters. Furthermore, we assume that the initial function
f is smooth and satisfies the boundary conditions, i.e. f(0)
=f(1) =0. Finally we assume that the mesh parameters h and
 are sufficiently small.
u mj
In order to compare
and
v mj
u ( x j , t m )   c e  t m sin(x j )
 1
  c e
  t m
sin(x j ) 
 1

 c e

  t m
 n 1
m
e  tm and (1  h ) .
2.2.2. Comparison of Fourier coefficient c and
coefficient  
We start by considering the Fourier coefficients, and note that
  is a good approximation of c because
, we note that

n
Since the sine part here is identical, it remains to compare the
Fourier coefficients c and   , and the time-dependent terms
sin(x j )
n
2  f ( x j ) sin(x j )
j 1
Here we want to show that

 c e 


is the trapezoidal-rule approximation of
sin(x j )  0
 tm
 n 1
1
(2.30)
Since f is smooth, it is also bounded and then the Fourier
coefficients c are bounded for all
Obviously, we have
.
2  f ( x) sin(x)dx
0
In fact we have
sin(x j )  1
and
2

  f (x
2
c e 



sin(x j ) 
 tm
 n 1

 c

 n 1
n
0
j 1
n
j 1

1
c     2 f ( x) sin(x) dx  2  f ( x j ) sin(x j )
e  t m sin(x j )
n
j 1

n
)  f ( x j ) sin(x j )  2  f ( x j ) sin(x j )
j 1


   f ( x j   )  f ( x j ) sin(x j )
j 1
1

e

 max c

C
 ( ) 2 t m
 n 1
 e 



2
t
 n 1
 ,t


t
m
2
t
n 1
 2t
  1  e
n1
2
t
 
 C e 
2
t
 2t
n2
 ...

2
1 e

2

f ( x j )  ... sin(x j )

e  tm , we simplify the problem a bit by choosing a fixed time
 2t
u mj   c e   t m sin(x j )
 1
tm , say tm=1 and we assume that
we want to compare the terms
2
2 . As a consequence,
(2.32)
and
1
(2.31)
v     (1  h  ) sin(x j )
m
 1
Motivated by the derivation of the solutions, we try to
compare the two sums term wise. Thus we keep  fixed, and
we want to compare
c e   tm sin(x j )
 (1  h ) m sin(x j )
and 
.
Unique Paper ID – IJTSRD27937 |
h
   e 
Now we want to compare the finite sums (2.26) and (2.31):
|
3
2.2.3
n
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f ( x j ) 
m
  t m
Comparison of the Terms (1  h ) and e
m
We will compare the term (1  h ) approximates the term
 e 2 t  ...
Since we have verified (2.30) it follow that
n

2
 O ( 2 ) , for f sufficiently smooth.
 0 , for large value of n.
m
j

n


3
    2 f ( x j ) 
f ( x j )  ...
2
j 1 

1
n 1
n
  
j 1
   e 
 C  e 

 C e 
n


2
    f ( x j )  f ( x j ) 
f ( x j )  ...  f ( x j ) sin(x j )
2
j 1 

   (1  h ) h
(2.33)
Since both   and  are very small for large values of  , it
is sufficient to compare them for small  . In order to
compare   and  for small values of  , we start by
recalling that
sin( y )  y  O ( y 3 )
Thus, we get
3
5


 h 2
 h 2


( )3  
( )5  



h
h
2
2


2


2 sin    2


 ...
2
2
3
!
5
!








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

2 h 
4 h 
 ( )   ( )  

h
2
2 
 2   ...
 2  1 

2
3!
5!




2
 2 
2
h
2
l mj  ut ( x j , t m )  utt ( x j , t m )  ...  u xx ( x j , t m )  u xxxx ( x j , t m )  ...
2
12
2
h
2
 u t ( x j , t m )  u xx ( x j , t m )  utt ( x j , t m )  u xxxx ( x j , t m )  ...
2
12
h
2
 utt ( x j , t m )  u xxxx ( x j , t m )  ...
2
12
h
2 , for h sufficiently small
  2 2 h
 O( 2 , h)
By using these facts, we derive
   (1  h )
 The finite difference scheme (2.6) is consistent of order
1
h
4
 

 1  h 2 sin 2 (
)
2 


(2,1).
Example 2.3.1
Solve the IVP
1
h
ut  u xx for x  (0,1) , t > 0
u (0, t )  u (1, t )  0 for t  0
1


h   h

 1  2 sin 2  


2



u ( x,0)  f ( x) for x  (0,1) .
The exact solution is given by
1
   (1   2 2 h) h
2
u  e  t sin x
11 
  1
1
1
h h 
2 2
 1  (  h)  
( 2 2 h) 2  ...  ( 2 2 h) h
h
2!
 1   2 2 
 e 
e
2
2
(2.34)
Approximate solution and exact solution are
illustrated by 2.1.
1
1 h
( 2  2 ) 2  ...  ( 2  2 h) h
2
, for h sufficiently small
 
  
This shows that also the time dependent term e
m
approximated by its discrete (1  h ) .
  t m
is well
2.3 Consistency
Lemma 2.3.1.
The finite difference scheme (2.6) is consistent of order (2,1).
Proof:
Local truncation error of the finite difference scheme (2.6) is
given by
u mj1  u mj
1
l mj 
 2 u mj1  2u mj  u mj1 
h

hl mj  u mj 1  u mj 
h
2
 u ( x j , t m )  hu t ( x j , t m ) 
u
m
j 1
 2u mj  u mj1

h2
h3
u tt ( x j , t m )  u ttt ( x j , t m )  ...
2
6


2
3
u ( x j , t m )  u x ( x j , t m )  2 u xx ( x j , t m )  6 u xxx ( x j , t m )



4
h  

 u ( x j , t m )  2  u xxxx ( x j , t m )  ...  2u ( x j , t m )  u ( x j , t m )  u x ( x j , t m )
  24

2
3
4
 


u
(
x
,
t
)

u
(
x
,
t
)

u
(
x
,
t
)

...
xx
j
m
xxx
j
m
xxxx
j
m
 2

6
24


 hut ( x j , t m ) 

h2
utt ( x j , t m )  ...
2

h  2
4
 u xx ( x j , t m )  u xxxx ( x j , t m )  ...
2 
 
12

@ IJTSRD
|
Unique Paper ID – IJTSRD27937 |
Figure2.1: Comparison of Exact and Approximate
Solutions
MATLAB codes for Figure 2.1
N=100 ; x=zeros (N,1) ;
g= zeros (N,1) ;
v=zeros (N,N) ;
%delta x=h, delta t=p
%f (x)=cos (2*pi*x) ;
H=1/(N+1) ;
P=1/(N+1) ;
for j = 1:N
x ( j )=j*h ;
end
%for m=1:N
% t ( m )=m*p ;
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International Journal of Trend in Scientific Research and Development (IJTSRD) @ www.ijtsrd.com eISSN: 2456-6470
%end
%gamma k
for k=1:N
sum=0 ;
for j=1:N
term=2*h*sin ( 2*pi*x ( j ))*sin (k*pi*x ( j ));
sum=sum+term ;
end
g ( k )=sum ;
end
%vk
for k=1 : N
for j = 1 : N
v ( k , j )=p*(4/h^2)*(sin(k*pi*h/2))^2)*sin(k*pi*x(j)) ;
%%%Approximate solution
plot ( x , s , ‘ – ‘) ; hold on
%%%%%%Exact solution
Plot ( x ,-exp (-*pi^2*0.01 )*sin(pi*x) ) ;
3. CONCLUSION
The aim of this research paper describe the Errors between
Analytical solution and Discretized solution of Differential
equations.
4. REFERENCES
[1] R. J. LeVeque, Finite difference methods for differential
equations, Washington, 2006.
end
v(k,:);
[2] J. W. Thomas, Numerical partial differential equation:
Finite difference methods, Springer-Verlag, New York,
1995.
end
sum 1 =0 ;
for k= 1 : N
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term1=g ( k ) . *v ( k , :) ;
sum1=sum1+term1 ;
s =sum1 ;
end
|
Unique Paper ID – IJTSRD27937 |
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