SKFKSD

1. Numerical solution of ordinary and
partial dierential Equations
Module 23: Analysis of

2. nite dierence
methods
Dr.rer.nat. Narni Nageswara Rao
£
August 2011
1 Convergence of dierence schemes
We are now establishing the convergence of the dierence scheme for the
numerical solution of the boundary value problem
yHH + f(t; y) = 0; a t b (1)
y(a) =
1; y(b) =
2
Consider the second order dierence scheme for (1) given by
yj 1 + 2yj yj+1 + h2
fj = 0; j = 1; 2; ¡¡¡ ; N (2)
The boundary conditions become
y0 =
1; yN+1 =
2 (3)
The exact solution y(t) of (1) satis

3. es
y(tj 1) + 2y(tj) y(tj+1) + h2
f(t; y(tj)) + Tj = 0 (4)
Where Tj is the truncation error. Substracting (4) from (2) and applying
the mean value theorem, and substituting j = yj y(tj), we get the error
equation
j 1 + 2j j+1 + h2
fyjj Tj = 0; j = 1; 2; ¡¡¡ ; N (5)
£nnrao maths@yahoo.co.in
1

4. where Tj =
h4
12
y(4)
(j); tj 1 j tj+1
In matrix notation, we write (5) as
ME = T (6)
where M = J + Q; E = [1; 2; ¡¡¡ ; N ]T
, T = [T1; T2; ¡¡¡ ; TN ]T
J =
2
6664
2 1 0
1 2 1
...
0 1 2
3
7775
and
Q = h2
2
6664
fy1 0
fy2
...
0 fyN
3
7775
From (6), we can observe that the convergence of the dierence scheme de-
pends on the properties of the matrix M. we noe show that the matrix
M = J +Q is an irreducible, monotone matrix such that M ! J and Q ! 0.
Irreducible: A tridiagonal matrix A = ai;j, where aij = 0 for ji jj 1
is irreducible if and only if ai;i 1 T= 0; i = 1; 2; ¡¡¡ ; N and ai;i+1 T= 0; i =
1; 2; ¡¡¡ ; N 1.
Diagonally dominant: A matrix A = ai;j is diagonally dominant if
jai;ij !
nX
j=1;iT=j
jai;jj; i = 1; 2; ¡¡¡ ; N
A matrix A = aij is said to be irreducibly diagonal dominant, if it is irre-
ducible and diagonally dominant with inequality being satis

5. ed for atleast
one i.
Theorem 1.1. A matrix A is monotone if Az ! 0 implies z ! 0.
The following are some properties of monotone matrices.
(i) A monotone matrix A is non singular.
(ii) A matrix A is monotone if and only if A
1
! 0.
2

6. Theorem 1.2. If a matrix A is irreducibly diagonal dominant and has non
positive o-diagonal elements, then A is monotone.
For example, the matrix J(see(6)) is tridiagonal, irreducibly diagonally dom-
inant and has non positive o diagonal elements. Therefore, J is a monotone
matrix.
Theorem 1.3. If matrices A and B are monotone and B A, then B
1
!
A
1
.
Now, consider equation (6).
Since fyj 0, j = 1; 2; ¡¡¡ ; N we have Q ! 0 and hence
M = J + Q ! J
Also, J is monotone. Now Q is diagonal matrix with positive diagonal en-
tries. Hence M = J + Q is also irreducibly diagonal dominant and have non
positive o-diagonal elements. Therefore, M is monotone and M ! J and
form Theorem 1.3 we have 0 M
1 J
1:
From (6), we have
E = M
1
T
kEk kM
1
kkTk kJ
1
kkTk (7)
In order to

7. nd this bound, we determine J
1
= ji;j explicitly. On multiplying
the rows of J by the jth
column of J
1
, we have the following equations.
(i) : 2ji;j j2;j = 0
(ii) : ji 1;j + 2ji;j ji+1;j = 0 i = 2; 3; ¡¡¡ ; j 1
(iii) : jj 1;j + 2jj;j jj+1;j = 1
(iv) : ji 1;j + 2ji;j ji+1;j = 0; i = j + 1; j + 2; ¡¡¡ ; N 1
(v) : jN 1;j + 2jN;j = 0 (8)
The solution of ((8).ii) using ((8).i), is given by
ji;j = c2i; i = 1; 2; 3; ¡¡¡ ; j 1 (9)
where c2 is independent of i, but may depend on j. Similarly, the solution of
((8).iv), using ((8).v), is given by
jij = c1
1 i
N + 1
; i = j + 1; j + 2; ¡¡¡ ; N 1 (10)
3

8. The constant c1 depends only on j. on equating the expression for ji;j ob-
tained from (9) and (10) for i = j we get
c2j = c1
1 i
N + 1
(11)
Also, on substituting the values of ji;j; i = j 1; j + 1 obtained from (9) and
(10) in ((8).iii), we have
c2 +
c1
N + 1
= 1 (12)
Finally, from (11) and (12), we get
c1 = j; c2 =
N j + 1
N + 1
(13)
on substituting the values of c1 and c2, we have
ji;j =
( i(N j+1)
(N+1)
; i j
j(N i+1)
(N+1)
; i ! j
(14)
From (14), we see that J
1
is symmetric.
The row sum of J
1
is gives as
NX
j=1
ji;j =
i(N i + 1)
2
=
(ti a)(b ti)
2h2
Hence, we obtain
kJ
1
k = max
1 i N
NX
j=1
jji;jj (b a)2
8h2
The equation (7) becomes
kEk (b a)2
8h2
kTk
Substituting, kTk h4M4
12
, we obtaine
kEk 1
96
(b a)2
h2
M4 = y(h2
) (15)
where M4 = maxjP[a;b] jy(4)
(j)j:
From the equation (15), it follows that the method is of second order and
kEk 3 0 or yj 3 y(tj) as h 3 0. This establishes the convergence of the
second order method.
4

9. 2 Stability of

10. nite dierence schemes
We now examine the stability of

11. nite dierence formulations. We take a
simple second order dierential equation with a signi

12. cant

13. rst derivative.
yHH + kyH = 0 (16)
where k is a constant such that jkj 1. Three dierent approximations
for (16) in which the

14. rst derivative is replaced by central, backward and
forward dierences respectively are
(i)
yj+1 2yj + yj 1
h2
+
k(yj+1 yj 1)
2h
= 0
(ii)
yj+1 2yj + yj 1
h2
+
k(yj yj 1)
h
= 0
(iii)
yj+1 2yj + yj 1
h2
+
k(yj+1 yj)
h
= 0 (17)
The analytical solution of (16) is given by
y(t) = A1 + B1e kt
where A1 and B1 are arbitrary constants to be determined with the help of
boundary conditions.
The characteristic equation corresponding to the dierence equation ((17).i)
is
1
h2
(2
2 + 1) +
k
2h
(2
1) = 0
2( 1)2
+ kh(2
1) = 0
( 1)[2( 1) + kh( + 1)] = 0
( 1)[(2 + hk) (2 h)] = 0
giving = 1, 2 hk
2+hk
) Solution of ((17).i) is
yj = A1 + B1
2 hk
2 + hk
j
= A1 + B1
1 kh
2
1 + kh
2
#j
5

15. If the behaviour of the exponential term is analysed, it is seen that it displays
the correct monotonic behaviour for k 0 and k 0 if the condition h 2
jkj
is satis

16. ed. This is the condition for the stability of the dierence equation
((17).i).
For very large k, 2 hk
2+hk
3 1 and due to the presence of the term ( 1)j
,
the solution (error) oscillates. Therefore, the stability condition will make
this central dierence scheme computationally infeasible.
Consider, the dierence equation ((17).ii). The characteristic equation
corresponding to it is
( 1)2
+ kh( 1) = 0
( 1)[ (1 hk)] = 0
) = 1; 2 = 1 hk
) yj = A1 + B1(1 hk)j
Analysis of the exponential term gives that if k 0, then we require that
jkhj 1 A h 1
k
for stability. If k 0, then there is no condition on h and
the dierence scheme ((17).ii) is unconditionally stable. If k becomes very
large and positive, then (1 hk) 3 I and due to the prescence of the
term ( 1)j
, the solution(error) oscillates. Therefore, a backward dierence
scheme becomes infeasible for large positive k.
Now consider the forward dierence scheme ((17).iii). The characteristic
equation is
( 1)2
+ kh(2
) = 0
( 1)[(1 + hk) 1] = 0
) 1 = 1; 2 =
1
1 + hk
) yj = A1 + B1
1
1 + kh
j
Again if k 0, then h 1
k
is the condition for stability.
If k 0, then there is no condition on h and proper behaviour is guar-
anteed for all h. Thus, if k becomes bery large and negative then a forward
dierence scheme is infeasible.
6

17. Hence, for stability it is necessary that dierent dierence approximations
for the

18. rst order term must be used depending on the sign of k. we may use
the approximation
yH(tj) =
yj yj 1
h
; if k 0
yj+1 yj
h
; if k 0
The one-sided dierence scheme is unconditionally stable and it is always on
the upstream or upwind side of tj. However, it suers from the disadvan-
tage that it is only

19. rst-order accurate.
3 Appendix
We recall the concept of a norm of a vector, kxk. The negative quantity kxk
is a measure of the size or length of a vector satisfying.
(i) kxk 0, for x T= 0 and k0k = 0
(ii) kcxk = jcjkxk, for any arbitrary complex number c.
(iii)
kx + yk kxk+ kyk (18)
We shall in most cases use the maximum norm
kxk = max
1 i n
jxij
At this point we must also recall the concept of a matrix norm. In addition
to the properties analogous to (18) the matrix norm must be consistent with
the vector norm that we are using for any vector x and matrix A.
kAxk kAkkxk
It is easy to verify that the norm
kAk = max
1 i n
nX
j=1
jai;jj
is consistent with max norm kxk.
7