1. The document discusses the convergence and stability of finite difference methods for solving ordinary and partial differential equations numerically. 2. It shows that the error of the finite difference solution converges to zero as the grid size decreases, establishing convergence of the method. 3. It analyzes the stability of explicit finite difference schemes for a test differential equation. The type of difference approximation used for the first derivative term depends on the sign of a constant, in order to satisfy the stability condition for the scheme.

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