# Rmk The symbol for is

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```					                                        vm  vm 1
Rmk: The symbol for Pk , h vm 
n
is
h
e skne imh  e skni ( m 1) h
h
(1  e ih ) skn imh
             e e
h
1  e ih           1  e  h
 p k , h ( s,  )            and   
h                   h
so it’s simply replacing  by i
sinh  sinh 
Rmk:  0                   
h        h
1
sinh h
2 (        2 ) 2  h  2 sinh 1 h
1                          2
h
2
1 h
sinh
1 h                   2
 h  2 sinh         
2             1
h
2
h
sinh(2 sinh 1 )
sinh h                      2 
 0           
h                     h
2 sinh 1
2
h
2 sinh 1
              2      
or                   h 0
sinh(2 sinh 1    )
2
h
1   sinh 1
h       
2 
 [1  (        )2 ] 2                   0
2                1
h
2
Ex: A(4.4) scheme for ut  au x  f
u m 2  u m 2
n        n
2k 2
 ut       uttt  O(k 4 )
4k                3
2k 2 2
 (1       t )ut  O(k 4 )
3
2k 2 2
 (1       t )(au x  f )  O(k 4 )
3
2k 2 2
 (1      t ) f
3
2k 2 2         h 2 2
 (1       t )a(1         ) 0 u  O(k 4 )  O(h 4 )
3                6
h 2 2         2u m1  u m  2u m1
n       n      n
 (a (1         ) 0 )(                       )
6                     3
2 f m 1  3 f m  2 f m 1
n          n       n
(                             )  O(k 4 )  O(h 4 )
3
Rmk: It’s stable for
3            1
| a |      (                     1
)
4                   3
(1  6 )( 6  )   2
2
Ex: Consider u (t n1/ 3 , x)
u n 1  u n               k n
 utn 1/ 3  utt 1/ 3  O(k 2 )
k                    6
k         n
 utn 1/ 3  (a 2u xx 1/ 3  f t n 1/ 3  af xn 1/ 3 )  O(k 2 )
6
n 1/ 3     n 1  2 n
and using                                O( k 2 )
3
we have

vm1  vm
n      n
vm1  2vm k 2 2 vm1  2vm
n       n          n     n
 a 0 (            ) a  (         )
k                   3       6        3
f m 1  f m ak
n        n
f m 1  2 f m
n          n
               0 (                )
2       6            3
the scheme is (2,2) scheme and is stable if | a | 3
Ex: From (3.2.2),

k t  vm  v n  v n1  (1  e k )v n               ,

(1  kt  )v  e v
n           kt n
,
 t v n 1  k 1n(1  k t  )v n 1
and by (3.2.9)

 t u n 2 / 3  e  k t / 3 t u n1
1
 (1  k t  )k 1n(1  k t  )u n 1  O(k 2 )
3
1               1
 (1  k t  )( t   k t2 )u n 1  O(k 2 )

3               2
1
 ( t   k t2 )u n 1  O(k 2 )

6
7u n 1  8u n  u n 1
                          O( k 2 )
6k
(2u x 1  u x )
n       n
and by u x  2 / 3 
n
 O( k 2 )
3
We have
7vm1  8vm  vm1
n      n    n
2vm1  vm
n      n
 a 0 (            )  f m 2 / 3
n
6k                      3
it’s shown in (Ex 4.3.1) that this scheme is
unconditionally stable.

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