1
Initial-Value Problems for ODEs
• numerical errors (round-off and truncation errors)
Consider a perturbed system:
• Does z(t) � y(t)?
(i) (uniqueness) a unique solution y(t) exists
� � � � � �
� �
GIVEN: , ,
FIND: for
ady y t f t y y a ydt
y t a t b
�� � �
� �
� � � �
� � 0
, ,
a
dz f t z t a t bdtz a y
� � � �
� �
2
� �� � � � �
� �
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� �
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0
0
0
(ii) (well-posed) for any >0, 0 such that
whenever and , with ,
a unique soltion to the problem
, ,
exists with
,
a
t C a b t
z tdz f t z t a t bdtz a y
z t y t a t b
� �
� � �
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� �
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�
� �
3
• sufficient conditions for the problem to be well posed:
� � � �
�
2 1 2 1
0
1 if , , , (Lipschitz condition);
2 if , and satisfies for some 0
f t y f t y L y y
ff C a b L Ly
� � �
�� � �
�
� � � � � �
� �
GIVEN: , ,
FIND: for
ady y t f t y y a ydt
y t a t b
�� � �
� �
4
� � Discretization: provide n ny y t� �
0 1 2
1
GIVEN: , , , ,
FIND:
n
n
y y y yy �
�
� � � � � �
� �
GIVEN: , ,
FIND: for
ady y t f t y y a ydt
y t a t b
�� � �
� �
5
• Classification:
� �� �
1 0 1 2
1 0 1 2 1
Explicit Schemes: , , , ,
Implicit Schemes: , , , , ,
n n
n n n
y F y y y yy F y y y y y
�
� �
���
��
��
� �� �
� �� �
1
1 1
1 1
1 1 1
one-step method:
or ,
multi-step method: , ,
or , , ,
n n
n n n
n n m n
n n m n n
y F yy F y yy F y yy F y y y
�
� �
� � �
� � � �
��� ���
��� ��
��
6
Taylor methods (explicit, one-step method)
• Taylor method of order k:
� � � �1
GIVEN: and ,
FIND:
n
n
y y t f t yy �
� �
� � � �
� � � � � � � � � � � �1 1
2 11 1
2 !
n n n
kk kn n n n
y y t y t dt
y t dt y t dt y t dt y t O dtk
� �
�
� � �
� ��� � � � � ��
7
� � � � � � � � � �need compute , , , ,k
n n n ny t y t y t y t� ��� �
� �n ny t y�
� � � �,n n n ny t f t y f� � �
� � � �,n n n
n nt t t t y y
d dfy t y t fdt dt� � �
� � � ��� � �� � �� � � �� � � �
� � � �2
2
,n n n
n nt t t t y y
d d fy t y t fdt dt� � �
� �� ���� �� ��� � �� � � �� � � �
� � � �� �notice that , ,f f t y f t y t� � �
� �� � � �, ,df d f dy ff t y t f f t ydt dt t dt y t y
� � � �� �� � � � � � �� � � �� �
8
� �2
2,
d f f f f t ydt t y t y
� � � �� �� �� � �� �� �� � � �� �� �
2
2f f f f f
t t y y t y y� �� � � � � � �� � � � � �� � � �� �� � � � � �� � � � � � �� � � � � �� �
22 2 22
2 22
f f f f ff f ft t y t y y y
� � � � � �� �� � � � �� �� � � � � � �� �
9
� �1
1truncation error per time step (from to ) kn nt t O dt �
�� !
accumulated truncation error from 0 to :t t T� � �
� � � �1 kkT dt O dt dtO T� �
� � � � � � � � � �2
1
1 1
2 !
kkn n n n ny y t dt y t dt y t dt y t
k� � ��� � � � ��
• Taylor method of order k:
10
• forward Euler method (k = 1, explicit, one-step):
� �1 ,n n n ny y dt f t y� � �
accumulated truncation error ~ O(T·dt)
simple but poor accuracy
� � � � 1i.e. , n nn n n
y yy t f t ydt
� �� � �
forward difference
• higher order (higher k): better accuracy but too much trouble
seldom in use
11
• backward Euler method (implicit, one-step):
� �1 1 1,n n n ny y dt f t y� � �� �
accumulated truncation error ~ O(T·dt)
� � � � 11 1 1, n n
n n ny yy t f t ydt
�� � �
�� � �
backward difference
12
� � � �� �1
1exact: ,n
n
t
n n n ty y t dt y f t y t dt�
� � � � � "
� � � � � �
� �
GIVEN: , ,
FIND: for
ady y t f t y y a ydt
y t a t b
�� � �
� �
� �,n ndt f t y � �1 1,n ndt f t y� �
13
• Crank-Nilcoson method (trapezoidal, implicit, one-step):
� � � �� �
� �1 1 1
2accumulated truncation err
, ,
or
2n n n n n n
dty y f t y f t y
O T dt
� � �� � �
�
• leap-frog method (explicit, two-step):
� �
� �
1
2
1
1 1 2 ,
if
accumulated truncation error
n n n
n n
n
n n
t t
y y dt f t y
T
t
dt
t dt
O
� �
� ��
�
�
�
� �
�
� � � � 1 1i.e. ,2
central difference
n nn n n
y yy t f t ydt
� ��� � �
�14
Runge-Kutta methods (explicit, one-step)
• What values of #, $, %1, %2 are required to obtain an accuracy of order 2?
2nd order method: accumulated truncation error ~ O( T·dt2 )
� �1 ,n nk dt f t y�
� �2 1,n nk dt f t dt y k$ #� � �
1 1 1 2 2n ny y k k% %� � � �
1 nk dt f�
� � � � � �2
2 1n n
nf fk dt f dt k O dtt y
$ #� �� �� � � �� �� �� �
� � � � � �2n nn n
f fdt f dt dt f O dtt y
$ #� �� �� � � �� �� �� �
� � � � � �� � � � � �2 2 2
2 22 3
1 12 2
1 1P.S.
2 2n n nf f fO dt dt dt k k O dtt t y y
$ $ # #� � �� � � �
� � � �
15
� � � � � � � �1 2
2
1n n
n n nn nf fdt f dt f dt dt f O dtt
y yy� � � �
� �� �� � �� �� �� �
% % $ #
� � � � � � � �2 2 3
1 2 2 2n n
n n nf fy dt f dt dt f O dtt y
% % $% #%� �� � � � � �
� �
On the other hand, the Taylor series expansion of yn+1 about tn is:
� �2 3
1
1
2n
n n ndfy y dt f dt O dtdt� � � � �
� �2 31
2n n
n n nf fy dt f dt f O dtt y
� �� �� � � � �� �� �� �
Thus, to have a scheme with a truncation error/step ~ O(dt3), we need
� �1 2
2 2
1
1
2
% %
#% $%
� �
� �
16
3
2
1 2 1truncation error
4 3 6
ff f f ft y
dty t y
� &� � � � �� � � �� �� �� � � �� '� �� � � �� � � � �� �� � � �� �� (#�
• 2nd Runge-Kutta method:
� �1 ,n nk dt f t y�
� �2 1,n nk dt f t dt y k$ #� � �
1 1 1 2 2n ny y k k% %� � � �
� �1 2
2 2
1
1
2
% %
#% $%
� �
� �
17
• 2nd Runge-Kutta method:
1 2
1
(1) 0, 1, 1 2 (midpoint method)
,2 2
n n n n ndt dty y dt f t y f�
� � � �
� �� � � �� �� �
% % # $
� �
1 2
1
(2) 1 2, 1 (midpoint Euler method)
,2 2
n n n n n ndt dty y f f t dt y dt f�
� � � �
� � � � �
% % # $
1 2
1
(3) 1 4, 3 4, 2 3 (Heun's method)
3 2 2,
4 4 3 3n n n n n n
dt dt dt dty y f f t y f�
� � � �
� �� � � � �� �� �
% % # $
� � � �1 1 2 1, ,n n n n n ny y dtf t y dtf t dt y k� � � � � �% % $ #
18
• 3rd Runge-Kutta method:
� �1 ,n nk dt f t y�
� �2 1 1 1,n nk dt f t dt y k$ #� � �
1 1 1 2 2 3 3n ny y k k k% % %� � � � �
� �3 2 2 1 3 2,n nk dt f t dt y k k$ # #� � � �
8 parameters!
6 constraints for O(dt4) per time step!
19
• 4th Runge-Kutta method (the most common one):
� �1 ,n nk dt f t y�
12 ,
2 2n ndt kk dt f t y� �� � �� �
� �
� � � �5
1 1 2 3 4
12 2
6n ny y k k k k O dt� � � � � � �
� �4 3,n nk dt f t dt y k� � �
23 ,
2 2n ndt kk dt f t y� �� � �� �
� �
� �4accumulated truncation error O T dt �
20
Adams-Bashforth methods of order m (explicit, multi-step)
� �,j j jf f t y�
� �1
1
1
0
mm
n n j n jj
y y dt b f O dt�
�� �
�
� � �)
� � � �1
0 1 1 1 1
mn n n m n my dt b f b f b f O dt �
� � � �� � � � � ��
e.g. Adams-Bashforth methods of order 3 (explicit, multi-step):
� �1 0 1 1 2 2n n n n ny y dt b f b f b f� � �� � � �
n-m+1 n-m+2 n-1 n n+1
21
� �
� �
0
2 3 4
1
2 3 4
2
1 1
2 6
42 2
3
n
n n n n n
n n n n
b f
y dt b f dt f dt f dt f O dt
b f dt f dt f dt f O dt
� &� �� �� �� �� �� ���� � � � � � �� '� �
� �� �� �� �� �� ���� � � � �� �� �� �� (
• Adams-Bashforth methods of order 3 (explicit, multi-step):
� �1 0 1 1 2 2n n n n ny y dt b f b f b f� � �� � � �
� � � � � � � �2 3 41 2
0 1 2 1 2
42
2n n n n
b by b b b dt f b b dt f dt f O dt
�� ��� � � � � � � � �
� �2 3 41 1
2 6n n n ny dt f dt f dt f O dt� ��� � � � �
22
Therefore,
0 1 2
1 2
1 2
1
2 1 2
4 1 3
b b bb bb b
� � ���� � ��� � ��
0
1
2
23 12
16 12
5 12
bbb
���* � ��� ��
� �1 1 223 16 512
n n n n ndty y f f f� � �� � � �
Conclusion:
� Adams-Bashforth three-step method
� accumulated truncation error ~ O(T·dt3)
23
Adams-Moulton methods of order m+1 (implicit, multi-step)
� �1
2
1
1
mm
n n j n jj
y y dt b f O dt�
�� �
�
� � �)
� � � �2
1 1 0 1 1 1 1
mn n n n m n my dt b f b f b f b f O dt �
� � � � � �� � � � � � ��
e.g. Adams-Moulton methods of order m+1=3 (implicit, multi-step):
� �1 1 1 0 1 1n n n n ny y dt b f b f b f� � � �� � � �
n-m+1 n-m+2 n-1 n n+1
24
• Adams-Moulton methods of order m+1=3 (implicit, multi-step):
� �
� �
2 3 4
1
0
2 3 4
1
1 1
2 6
1 1
2 6
n n n n
n n
n n n n
b f dt f dt f dt f O dt
y dt b f
b f dt f dt f dt f O dt
�� &� �� �� ���� � � �� �� �� �� �
� � �� '� �� �� �� ���� �� � � � �� �
� �� (
� � � � � � � �2 3 41 1
1 0 1 1 12
n n n nb b
y b b b dt f b b dt f dt f O dt�� �
�� ��� � � � � � � �
� �2 3 41 1
2 6n n n ny dt f dt f dt f O dt� ��� � � � �
� �1 1 1 0 1 1n n n n ny y dt b f b f b f� � � �� � � �
25
Therefore,
1 0 1
1 1
1 1
1
1 2
1 3
b b bb bb b
�
�
�
� � ��� � ��� � ��
1
0
1
5 12
8 12
1 12
bbb
� ���* ��� � ��
� �1 1 15 812
n n n n ndty y f f f� � �� � � �
Conclusion:
� Adams-Moulton two-step method
� accumulated truncation error ~ O(T·dt3)
26
predict-corrector methods
accumulated truncation error ~ O(T·dtm+1)
� �1
* 1
1
0
mm
n n j n jj
y y dt b f O dt�
�� �
�
� � �)
*
1PREDICTOR: use Adams-Bashforth -step method to predict :nm y �
� �* *
1 1 1,n n nf f t y� � ��
1CORRECTOR: use Adams-Moulton -step method to predict :nm y �
� �1
* 2
1 1 1
0
mm
n n n j n jj
y y dt b f b f O dt�
�� � � �
�
� �� � � �� �
� �)
advantage: explicit
disadvantage: need many more computations
27
2 31 1 (expected)
2 6n n n ny dt f dt f dt f� �� ���� � � � ��
1
1
0 explicit schemes
0 implicit schemes
b
b
�
�
���� �+��
� �1 0 1 1 1 1
1 1 0 1 1 1 1
n n n m n m
n n n m n m
y a y a y a y
dt b f b f b f b f
� � � � �
� � � � � �
� � � �
� � � � �
�
�
General multi-step methods:
0AB/AM methods: 1, 0 for 1ja a j� � � ,
28
2 explicit schemesdegrees of freedom (# of adjustable variables):
2 1 implicit schemes
mm
�� � ��
2 1 explicit schemesmaximum order of accuracy attainable:
2 implicit schemes
mm
��� �
�
General multi-step methods:
2 31 1 (expected)
2 6n n n ny dt f dt f dt f� �� ���� � � � ��
� �1 0 1 1 1 1
1 1 0 1 1 1 1
n n n m n m
n n n m n m
y a y a y a y
dt b f b f b f b f
� � � � �
� � � � � �
� � � �
� � � � �
�
�
29
� �
� �
� �
11 1 2
22 1 2
1 2
, , , ,
, , , ,
, , , ,
N
N
NN N
du f t u u udtdu f t u u udt
du f t u u udt
� ���� ������ ���
�
�
�
�
Systems of differential equations
� � � � � �1 1 2 2 NI.C.s: 0 = , 0 = , , 0 =Nu u u# # #�
� �� �
� �
1 1 21
2 1 22
1 2
, , , ,
, , , ,
, , , ,
N
N
N N N
f t u u uuf t u u uu
ddt
u f t u u u
� �� �� �� �� �� �� �� � �� �� �� �� �
� � � �� � � �
��
� �� �
�
30
� �
� � � �1 2
,
0 , , ,T
N
dU F t UdtU # # #
�
� �
� � � �1 2 1 2Let , , , and , , , , ThenT T
N NU u u u F f f f� � �� �
� �1 1 15 812
n n n n ndty y f f f� � �� � � �
e.g. Adams-Moulton two-step method (3rd order)
1 1 1 1 1
2 2 2 2 2
1 1 1
5 812
N N N N Nn n n n n
u u f f fu u f f fdt
u u f f f� � �
� &� � � � � � � � � �� �� � � � � � � � � �� �� � � � � � � � � �� � � �� '� � � � � � � � � �� �� � � � � � � � � �� �� � � � � � � � � �� (
� � � � �
31
� �� �
� �� �� �
1 1 1
1 1 1 2
1 1 1
2 1 1 2
1 1 1
1 1 2
11 1 2
1 1
12 1 22 2
1
1
, , , ,
, , , ,5
, , , ,
, , , ,
, , , ,8
12
n n nn N
n n nn N
n n nN n N
n n nn nn N
n n nn nn N
n nN Nn N n
f t u u u
f t u u u
f t u u u
f t u u uu uf t u u uu u dt
u u f t
� � ��
� � ��
� � ��
�
�
��
� �� �� �� �� �� �� �
� � � �� � � �� � � �� � �� � � �� � � �� � � �� � � �
�
�
��
�
�� � �
� �� �� �
� �
1 2
1 1 1
1 1 1 2
1 1 1
2 1 1 2
1 1 1
1 1 2
, , , ,
, , , ,
, , , ,
, , , ,
n n nN
n n nn N
n n nn N
n n nN n N
u u u
f t u u u
f t u u u
f t u u u
� � ��
� � ��
� � ��
� &� �� �� �� �� �� �� �� �� �� �� �� �� �� '� �� �� �� �� �
� �� �� �� �� �� �� �� �� ��� �� �� �� �� �� �� �� (
�
�
�
��
1
2
1N n
ff
f�
� �� �� �� �� �� �
�
1
2
N n
ff
f
� �� �� �� �� �� �
�
1
2
1N n
ff
f�
� �� �� �� �� �� �
�
32
2 1
2 1, , , ,
m m
m md y dy d y d yf t ydt dt dt dt
�
�
� �� � �
� ��Higher order equations
� � � � � � � �2 1
1 2 3 m2I.C.s 0 = , 0 = , 0 = , , 0 =
m
mdy d y d yydt dt dt
# # # #�
�
� � � �
� �
� �
� �
1
12
2
23 2
1
1
Let
mm
m m
u t y tdu dyu tdt dtdu d yu tdt dt
du d yu tdt dt
��
���� � ���� � ������ � ���
�
�
� �
� �
� �
� �
12
23
1
1 2, , , ,
mm
mm
du u tdtdu u tdt
du u tdtdu f t u u udt
�
� ���� ������ ���
���
�
�
� � � � � �1 1 2 2 mI.C.s: 0 = , 0 = , , 0 =mu u u# # #�
33
Consistency: difference equation - differential equation
as dt- 0 ? (truncation errors - 0 as dt- 0)
Stability: computed solution to the difference equation
- exact solution to the difference equation?
(round-off errors under control)
Convergence: computed solution to the difference equation
- exact solution to the differential equation
as dt- 0 ?
34
� � � � � �� � 0
,
0
r i
y t f t y y ty y
i C
� � � .
�
. � . � . �
test problem
the exact solution is:
� � � � � � � �� �0 exp cos sinr i iy t y t t i t� . . � .
� �where is bounded for all 0 as long as 0.ry t t , . �
35
Schemes stability
(1) forward Euler method:
� �1 1 orn n n n n ny y dt f y dt y dt y. .� � � � � � �
� � � � � �1 1n nf y dt f y.� � �� �
� �� �1 1 1n n n ny e dt y e.� �� � � �
Alternatively, the solution of the difference equation is
� � 01 if 1 1 even if 0.n
n ry dt y dt. . .� � - / � � �
(numerical instability instead of physical instability)
� � � � � �,y t f t y y t� � � .
� �Suppose + , where is the round-off error. Thenn n n nf y y e e��
� �1Th s 1u n ne dt e� � � .
� � 0 1 if 1 1.n
ne dt e dt* � � . - / � . �
36
~ a limitation on the magnitude of dtbesides the consideration of accuracy
(1) forward Euler method
stability criterion: 1 1dt� . �
Numerical Stability
1
37
(2) backward Euler method:
1 1 1n n n n ny y dt f y dt y� � �� � � � .
� � � �1
1 01 1n
n n ny dt y y dt y. .� �� � � * � �
stable if 1 1.dt* � ,.
Inconsistent with the exact solution
when 0 and 1 1!rdt dt. � � . ,
1because the Taylor's secies of has a convergence radius of 1.
1 z�38
backward Euler method
2 31 11 for (1)
2! 3!
ze z z z z� � � � � � /�
211 for 1 (2)
1z z z
z� � � � �
��
Regions I, II, and V: truncation errors under control.
Regions I, and III: rounding errors out of control.
39
0 Region I: Both series converge to a value > 1
Rounding error diverges.
40
0 Region II: Exact series converge to a value > 1
Numerical series converge to a value < 1.
Rounding error converges.
1
41
0 Region III: Exact series converges to a value > 1.
Numerical series does not converge.
Rounding error diverges.
42
0 Region IV: Exact series converges to a value > 1.
Numerical series does not converge.
Rounding error converges.
43
0 Region V: Both series converge to a value < 1.
Rounding error converges.
44
0 Region VI: Exact series converges to a value < 1.
Numerical series does not converge.
Rounding error converges.
1
45
(3) Trapezoidal (Crank-Nilcoson) method:
� � � �1 1 1
1 1
2 2n n n n n n ny y dt f f y dt y y.� � �� � � � � �
1
12
12
n n
dt
y ydt�
� ��� �� � �
�� �� �
.
. 0
12
12
n
n
dt
y ydt
� ��� �� � �
�� �� �
.
.
stable if 12
0 12
rdt dtdt
� � � �1 .. .
46
(4) leapfrog method:
1 1 12 2n n n n ny y dt f y dt y.� � �� � � �
Suppose . Substitute into the difference equation to oftainnny 2�
2 2
1,2
2 2 1 0 1tdt dt d* �� � 3 �� 2. . .2 2
1 1 2 2 1,2stable if 1n nny C C2 2 2� � * �
1 2 1,2However, since 1 stable if 1 � � * �2 2 2
stable 0 if 1r idt and dt. � . �*
� � � �1 2Write exp . Then exp . i i� � � �2 4 2 4
1 2Moreover, 2 2 sindt i� � �2 2 . 4
47
(5) Runge-Kutta methods:
� �212nd method: 1+ 1
2dt dt. .� �
� � � �2 31 13rd method: 1+ 1
2 6dt dt dt. . .� � �
� � � � � �2 3 41 1 14nd method: 1+ 1
2 6 24dt dt dt dt. . . .� � � �
48
• 2nd order Runge-Kutta methods:
� �1 2
2 2
1
1
2
� �
� �
% %
#% $%
1 n nk dt f dt y� � .
� � � � � �2 1 1,n n n n nk dt f t dt y k dt y k dt y dt y� � $ � # � . � # � . � #.
� �
� � � �� �1 1 1 2 2 1 2
2
1 2 21
n n n n n n
n
y y k k y dt y dt y dt y
dt dt y
� � � % � % � � % . � % . � #.
� � % � % . � #% .
� �5 6211
2ndt dt y� � . � .
� �21stable if 1 1
2dt dt� . � . �
1
49
• In general:
� �1 0 1 1 1 1
1 1 0 1 1 1 1
n n n m n m
n n n m n m
y a y a y a y
dt b f b f b f b f
� � � � �
� � � � � �
� � � �
� � � � �
�
�
� �� �
1 2
0 1 1
1
1 0 1
Define
m m mm
m mm
p z z a z a z a
q z b z b z b
� ��
�� �
� � � � �
� � � �
�����
�
�
• The multi-step method is said to be stable if all roofs of p(z)lie in the disk |z|�1 and if each root of modulus 1 is simple.
• The method is said to be consistent if � � � � � �1 0 and 1 1p p q�� �
Theorem For the multi-step method to be convergent, it is
necessary and sufficient that it be stable and consistent.
50
Conclusion:
implicit schemes: more stable – allow a larger time increment dt
troublesome
explicit schemes: less stable – need a small dt
easy to implement.
51
system of equations:
test problem: dU AUdt
�
where A is a constant complex matrix.
5 61
suppose are the complex eigenvalues of the matrix N
k k A�
.
stable if all stable regiondt. ��
52
Boundary-Value Problems for ODEs
� �
� � � �
, , ,
,
y g x y y a x b
y a y b
�� �� � �
� �# $
~ shooting method and finite difference method
1
53
shooting method 7 IV ODEs� �
� � � �, , ,
,
y g x y y a x by a y b# $
�� �� � �
� �
� �STEP1: guess and integrate the ODEs until :y a t x b� � �
� � � �
� �
� � � �
1
12
21 2
1 2
, ,
,
u x y x
du udx
du g x u udx
u a u a t
�
�
�
� �#
� � � �STEP2: check if the relative error ? if not, re-guess .y b
y a� $
�� $
� �y a t� �
54
• How to re-guess? Notice � � � �fn. of bty b y t� �
� � � �Define .bf t y t� � $
� � � �1 ,STEP2: if , then
b i iy t u b t� �� �
$ $
$ $
root-searching mehods. e.g., Secant method:*
� � � �0 1STEP1: take two initial guesses = and = .y a t y a t� �
� �Thus we are looking for a value of such that =0t f t
� �� �� � � �
1
1
1
i i ii
i ii
f t t tt
f f tt
t�
��
�� �
�
� �� �� �� � � �
1
1
b i i ii
b i b i
y t t tt
y t y t�
�
� $ �� �
�� �� �� �� � � �
1 1
1 1 1
,
, ,
i i ii
i i
u b t t tt
u b t u b t�
�
� $ �� �
�
55
• A special case: the ODE is linear
� � � � � � � �� � � �
, ,y g x y y p x y q x y r xy a y b
�� � �� � � �
� �# $
� � � � � �� �� �
� �1(1)
0
y p x y q x y r xy a y xy a
�� �� � �� &� �� *� '� �� �� (
#
� � � �� �� �
� �2(2) 0
1
y p x y q x yy a y xy a
�� �� �� &� �� *� '� �� �� ( � � � � � �
� �� �1
1 2
2
Theny b
y x y x y xy b�
� � $
56
Finite-difference methods
� � � � � �� �, ,i i i iy x g x y x y x�� ��
0BC's: , Ny y# $� �
implicit equations for , 1,2,3, , 1iy i N� �� �
root-searching for multi-variable system�
� �� � � �
, , ,
,
y g x y y a x by a y b# $
�� �� � �
� �
truncation error ~ O(h2)
1 1 1 1
2
2, ,
2
i i i i ii i
y y y y yg x yh h
� � � �� � �� �� � �� �
e.g. central difference:
for i = 1,2,…, N�1
1
57
� � � � � � � �linear equation: , ,g x y y p x y q x y r x� �� � �
� � � � � �1 1 1 1
2
2
2
i i i i ii i i i
y y y y yp x q x y r xh h
� � � �� � �� � �
1 12 2 2
1 2 1
2 2
i ii i i i i
p py q y y rh h h h h� �
� � � �� �� � � � � �� �� � � �� �� � � �
0
1 112 2 2
1 1
2 22 2 22 2 2
1 1 1 112 2 2
1 0 0 0
1 2 10 0
2 2
1 2 10 0 0
2 2
1 2 10 0
2 2
0 0 1
N N N NN
N
yp pq y rh h h h h
p pq y rh h h h h
p p y rqh h h h h
y
� � � ��
� �� � �� �� � �� �� �� � � � � � �� �
� �� �� � �� �� � �� �� �� � � � � � �� � �� �� �� �� �� �� �� �
� �� �� �� � � �� �� �� �� �� �� � �� �� �
#
$
� � �
� �
�� ��
�
�
������
� �� �� �� �� �� �
�