CE 601: Numerical Methods
Lecture 27
Multi-Point Methods
Course Coordinator:Dr. Suresh A. Kartha,Associate Professor,
Department of Civil Engineering,IIT Guwahati.
• Multi Point Methods:
• The Euler’s methods and Runge-Kuttamethods that were discussed till now for solving IV-ODEs were single step (or single point) methods.
• In the time scale, only the information of was considered for obtaining
ny
1ny
• If the information of more than one previous instant is used, then the method is called a multi-point method.
• Principle of Multi-Point methods:
• We can approximately represent the above equation as
( , )dy
f t ydt
( , ( )) ( )dy f t y t dt F t dt
( )dy F t dt
• at different instants is approximated by Newton’s backward difference polynomials.
• So, for the discrete time domain, we have
( )F t
( )kdy P t dt
1 1
1 1
( )n n
n q n q
y t
k
y t
dy P t dt
• If is obtained with base point time , the resulting expression will be explicit multi-step equation.
• If is obtained with base point time , then the resulting expression will be implicit multi-step equation.
( )kP t nt
( )kP t1nt
• If , we have
• This integration (1) is called Adams’ finite-difference equations.
• Explicit ones are called Adams-Bashforth FDEs.
• Implicit ones are called Adams-Moulton FDEs.
1q1 1
( ) (1)n n
n n
y t
k
y t
dy P t dt y
• Fourth Order Adams Bashforth Moulton Method:
• It has already been seen that
• In this case, a third degree polynomial is integrated to obtain the fourth degree polynomial. So,
1 1
( )n n
n n
y t
k
y t
dy P t dt y
1
3 ( )n
n
t
nt
y P t dt
• The explicit solution has been considered first.2 3 4
3
1
1
1 3
0
1
2 3
1
0
1
(s) . ( 1) / 2! ( 1)(s 2) / 3! ,
when
, 0
, 1
So, ( )
i.e. . ( 1) / 2! ( 1)(s 2) / 3!
i.e.
n n n n
n
n
n
n n
n n n n n n
n n
P f s f s s f s s f O t
t t dts ds
t t
t t s
t t s
y y y t P s ds
y y t f s f s s f s s f ds
y y t f 2 31 5 3
2 12 8n n n nf f f
• Backward difference table is as follows:2 3
3 3
n n
t f f f f
t f
2 3
2 2 1 2 3
1 2
2.
n n
n n n n n
n n
f f
t f f f f
f f 1 2 3
1 1 1 2
1
3. 3.
2.
n n n n
n n n n n
n n
f f f f
t f f f f
f f 1 1 2
1 1
1
1
3. 3.
2.
n n n n
n n n n n
n n
n
f f f f
t f f f f
f f
t 1 nf
• So, substituting all the relevant values in the equation, we have
Or,
This is called Adams-Bashforth fourth order
explicit FDE.
1 1 1 2 1 2 3
1 5 32. 3. 3.
2 12 8n n n n n n n n n n n ny y t f f f f f f f f f f
1 1 2 355 59 37 924
n n n n n n
ty y f f f f
• Similarly, in case we integrate using implicit approach,
• This is the fourth order Adams-Moulton implicit FDE.
• As it is observed difficulty in evaluating ‘f’ in implicit condition, so Adams-Moulton FDE can be simplified by predictor-corrector approach (known as 4th order Adams-Bashforth-Moulton predictor corrector
method):
1 0
3 1
1
1 1 1 2
( ) ( ) error
9 19 5 924
n
n
t
nt
n n n n n n
y P t dt P s ds
ty y f f f f
1 1 2 3
1 1 1 2
55 59 37 924
9 19 5 924
P
n n n n n n
C P
n n n n n n
ty y f f f f
ty y f f f f
• To solve IV-ODE’s having Non-Linear DerivativeFunctions
• For the IV-ODE
it has been seen that explicit and implicit
methods can be employed. However, when
is non-linear in , then employing the
implicit methods become quite tedious and
difficult.
0 0( , ); ( )dy
f t y y t ydt
( , )f t y y
• Let us consider the Euler implicit method
• Here, is non-linear. So, in the time scale
is unknown in
Let us write it as
1 1 1. ,n n n ny y t f t y
1 1,n nf t y
1 1. ,n n ny t f t y 1ny
1( )nG y
• Hence, we will need to solve the equation
• We need to find out the value of using iterative procedures. Let us define
• Newton Raphson method suggests that
1 1( ) 0n ny G y
1ny
1 1 1( ) ( )n n nF y y G y
( )
1( 1) ( )
1 1 ( )
1'
k
nk k
n n k
n
F yy y
F y
• Modified Newton Raphson method suggests that
• Here, is calculated only once.
( )
1( 1) ( )
1 1'
k
nk k
n n
F yy y
F
'F
• Example: Solve
using Euler’s implicit method and apply
Newton Raphson method.
• Solution:
4 4 12
0; 250 , 2500 , 4 10a a
dTT T T K T K
dt
1 1
12 4 4
2
where
4 10 250
n n n n
tT T f f
f T
• Using , we have
• Initial guess is taken as
2.0t s
12 4 4 4 4
1 0 1
12 4
1 1 1
12 4
1 1 1
11 3
1 1
2.04.0 10 2500 250 250
2
2343.78 4 10
2343.78 4 10
' 1 1.6 10
T T T
T T G T
F T T T
F T T
(0)
1 2500T
(0)
1(1) (0)
1 1 (0)
1'
F TT T
F T
• Eventually, after several iterations, we arrive at the answer as .
(1)
1
(1)
1(2) (1)
1 1 (1)
1
(2)
1
(3)
1
(4)
1
(5)
1
312.47002500 2083.373
0.7500
'
185.0492083.373 2299.725
0.8553
67.8282299.725 2215.508
0.8054
31.8992215.508 2254.127
0.826
13.6162254.127 223
0.8167
T
F TT T
F T
T
T
T
T 7.454
1 2242.605T K