1. Thurs 1/6
1
1. IVP for ODEs
y ! = f (y) ,
y(0) = y0
:
find y(t)
ex
y! = y
,
y ! = sin y
y(0) = y0
,
⇒
y(0) = y0
y(t) = y0 et
⇒
y(t) = ?
def
An IVP is well-posed if the following conditions are satisfied.
1. A solution exists.
2. The solution is unique.
3. The solution depends continuously on the data (i.e. y0 , f ).
def
f satisfies a Lipschitz condition on a domain D if there exists a constant L st
|f (y) − f (u)| ≤ L |y − u| for all y , u ∈ D.
note : we typically assume that f (y) is smooth, so L = max |fy |
thm
If f satisfies a Lipschitz condition, then the IVP is well-posed.
ex
y ! = ay + b , y(0) = y0 ⇒ L = |a| , y(t) = y0 eat +
b at
(e − 1)
a
y ! = y 1/2 , y(0) = 0 ⇒ L = ∞ for y ≥ 0 , non-unique solution , y(t) = 0 ,
y ! = y 2 , y(0) = 1 ⇒ L = ∞ for y ≥ 1 , y(t) =
1
, lim y(t) = ∞
1 − t t→1
t2
4
2
Euler’s method
h = ∆t = step size , tn = nh
yn = y(tn ) : exact solution
,
un : numerical approximation
y ! = f (y) : differential equation
un+1 − un
= f (un ) : finite-difference scheme
h
un+1 = un + hf (un ) , u0 = y0
ex
y ! = y , y0 = 1
u0 = 1
u1 = u0 + hf (u0 ) = 1 + h
u2 = u1 + hf (u1 ) = (1 + h) + h(1 + h) = (1 + h)2
···
un = (1 + h)n
tn = 1 ⇒ nh = 1 , yn = y(1) = e = 2.7182818 . . .
n
10
20
40
80
h
0.1
0.05
0.025
0.0125
un
yn − un (yn − un )/h
2.5937425 0.1245 1.245
2.6532977 0.0650 1.300
2.6850638 0.0332 1.328
2.7014849 0.0168 1.344
3
Euler’s method
y ! = y , y(0) = 1
un+1 = un + hun , u0 = 1
h=1
h=1/2
8
8
7
7
6
6
5
5
4
4
3
3
2
2
1
0
0.5
1
1.5
2
1
0
0.5
h=1/4
8
7
7
6
6
5
5
4
4
3
3
2
2
0
0.5
1
1.5
2
1.5
2
h=1/8
8
1
1
1.5
2
1
0
0.5
note
1. The solid line is the exact solution y(t).
The circles are the numerical solution un .
The dashed line is the piecewise linear interpolant of un .
2. For a fixed time t, the error decreases as h → 0.
3. For a fixed stepsize h, the error increases as t → ∞.
1
4
convergence proof (special case)
y ! = y , y0 = 1 ⇒ y(t) = et
un+1 = un + hun , u0 = 1 ⇒ un = (1 + h)n
1
n
consider t = 1 , h =
!
1
lim un = n→∞
lim 1 +
h→0
n
!
1
lim
n
ln
1
+
n→∞
n
"
"n
!
1
= n→∞
lim exp n ln 1 +
n
= ∞ · 0 = n→∞
lim
⇒ lim un = e = y(1)
h→0
!
⇒
ln (1 + n1 )
1
n
""
= n→∞
lim
!
1
= exp n→∞
lim n ln 1 +
n
1
1+
1
n
Euler’s method converges
hw : y ! = ay + b , y(0) = y0
convergence proof (general case)
y ! = f (y) , y(0) = y0
un+1 = un + hf (un ) , u0 : given
fix t > 0 , set h = t/n ⇒ t = nh = tn , yn = y(tn )
goal : lim un = yn
h→0
yn+1 = yn + hf (yn ) + τn
,
τn : local truncation error
step 1 : estimate τn
yn+1 = y(tn+1 ) = y(tn + h) = y(tn ) + hy ! (tn ) +
h2 !! #
y (t)
2
h2 !! #
h2 !! #
yn+1 = yn + hf (yn ) +
y ( t ) ⇒ τn =
y (t)
2
2
step 2 : analyze error , en = yn − un : error
en+1 = yn+1 − un+1 = yn + hf (yn ) + τn − (un + hf (un ))
= yn − un + h(f (yn ) − f (un )) + τn
= en + h(f (yn ) − f (un )) + τn
!
·
−1
n2
·
1
−1
n2
""
= 1
5
|en+1 | ≤ |en | + h|f (yn ) − f (un )| + |τn | ≤ |en | + hL|yn − un | + τ , τ = max |τn |
|en+1 | ≤ (1 + hL)|en | + τ
|e1 | ≤ (1 + hL)|e0 | + τ
|e2 | ≤ (1 + hL)|e1 | + τ ≤ (1 + hL)2 |e0 | + (1 + (1 + hL))τ
|e3 | ≤ (1 + hL)|e2 | + τ ≤ (1 + hL)3 |e0 | + (1 + (1 + hL) + (1 + hL)2 )τ
···
|en | ≤ (1 + hL)n |e0 | +
n−1
$
(1 + hL)j τ
j=0
(1 + hL)n − 1
(1 + hL)n − 1
(1 + hL) =
=
(1 + hL) − 1
hL
j=0
n−1
$
j
note : 1 + x ≤ ex ⇒ (1 + x)n ≤ enx ⇒ (1 + hL)n ≤ enhL = eLt
eLt − 1
τ
|en | ≤ e |e0 | +
hL
Lt
τ = max |τn | =
h2
M , M = max |y !! (t)|
2
eLt − 1 M h
|en | ≤ e |e0 | +
L
2
Lt
if lim |e0 | = 0 , then lim |en | = 0 ⇒ lim un = yn , i.e. Euler’s method converges
h→0
h→0
h→0
note
There are two types of convergence.
1. Fix t, set h = t/n. Does lim un = y(t)?
h→0
2. Fix h. Does n→∞
lim un = y(∞)?
2. Tues 1/11
6
roundoff error
un+1 = un + hf (un ) : Euler’s method in exact arithmetic
vn+1 = vn + hf (vn ) + "n : computed solution in finite precision arithmetic
yn+1 = yn + hf (yn ) + τn : exact solution
let en = yn − vn
en+1 = en + h(f (yn ) − f (vn )) + τn − "n
|en+1 | ≤ (1 + hL)|en | + τ + " , τ = max |τn | , " = max |"n |
eLt − 1
|en | ≤ e |e0 | +
L
"
→ ∞ as h → 0
h
Lt
!
Mh
"
+
2
h
"
note
eLt − 1 M h
|un − yn | ≤
: error bound
L
2
We may rewrite this as un − yn = O(h).
%
%
%
%
un − yn %%%
This means there exists a constant C, independent of h, st
% ≤ C for
h
eLt − 1 M
all h sufficiently small. We may take C =
.
L
2
ex
e−1 e
y ! = y , y(0) = 1 , L = 1 , t = 1 , M = e , C =
∼ 2.34
1 2
%
%
% un − y n %
%
% ∼ 1.34 as h → 0.
However, we saw computationally that %
%
h
claim
un − yn = hEn + O(h2 ) : asymptotic expansion
(Math 557)
%
%
% u − (y + hE ) %
n
n %
% n
% ≤C
This means there exists a constant C, independent of h, st %%
%
h2
for all h sufficiently small. The constant En is defined by En = E(tn ), where
E ! = fy (y)E − 12 y !! , E(0) = 0. E(t) is the principal error function.
7
ex
y ! = y , y(0) = 1
E! = E −
1
2
et ⇒ E(t) = −
1
2
⇒ un − yn = −1.36h + O(h2 )
tet , E(1) = −1.36
pf
define dn = un − (yn + hEn )
we already know that dn = O(h) , we must show that dn = O(h2 )
dn+1 = un+1 − (yn+1 + hEn+1 )
= un + hf (un ) − (yn + hyn! + 12 h2 yn!! + O(h3 ) + h(En + hEn! + O(h2 )))
= un − (yn + hEn ) + h(f (un ) − yn! ) − h2 ( 12 yn!! + En! ) + O(h3 )
f (un ) = f (yn + hEn + dn ) = f (yn ) + fy (yn )(hEn + dn ) + O(h2 )
f (un ) − yn! = fy (yn )(hEn + dn ) + O(h2 )
dn+1 = dn + h(fy (yn )(hEn + dn ) + O(h2 )) − h2 ( 12 yn!! + En! ) + O(h3 )
= dn + hfy (yn )dn + h2 (fy (yn )En − 12 yn!! − En! ) + O(h3 )
&
'(
dn+1 = dn + hfy (yn )dn + O(h3 )
)
= 0 by definition of E(t)
|dn+1 | ≤ (1 + hL)|dn | + κ , κ = O(h3 ) , d0 = 0
···
eLt − 1
|dn | ≤
κ = O(h2 )
hL
note
ok
un − yn = O(h)
un − yn = hEn + O(h2 )
un − yn = hEn + h2 Dn + O(h2 ) , where Dn = D(tn )
hw : find the equation satisfied by D(t)
8
application : Richardson extrapolation
un = yn + hEn + h2 Dn + · · ·
A00 = uh = y + hE + h2 D + O(h3 )
2
A10 = uh/2 = y +
h
2E
+ ( h2 ) D + O(h3 )
A20 = uh/4 = y +
h
4E
+ ( h4 ) D + O(h3 )
2
eliminate O(h) term
2A10 − A00 = A11 = y −
h2
2 D
+ O(h3 )
2A20 − A10 = A21 = y −
h2
8 D
+ O(h3 )
eliminate O(h2 ) term
4A21 − A11
= A22 = y + O(h3 )
3
note
Aj0 = y + O(h)
Aj1 = 2Aj0 − Aj−1,0 = y + O(h2 )
4Aj1 − Aj−1,1
= y + O(h3 )
3
8Aj2 − Aj−1,2
=
= y + O(h4 )
7
Aj2 =
Aj3
ex : y ! = y , y(0) = 1 ⇒ y(1) = e = 2.7182818
j
0
1
2
3
h
0.1
0.05
0.025
0.0125
Aj0
2.5937425
2.6532977
2.6850638
2.7014849
Aj1
2.7128529
2.7168299
2.7179060
O(h)
O(h2 )
Aj2
Aj3
2.7181556
2.7182647 2.7182803
O(h3 )
O(h4 )
down a column : decreasing step size , fixed order of accuracy , h-refinement
across a row : fixed step size , increasing order of accuracy , p-refinement
3. Thurs 1/13
9
Taylor series methods
y ! = f (y)
yn+1 = yn + hyn! +
h2
2
yn!! + · · ·
1st order Taylor series method
yn! = f (yn )
un+1 = un + hf (un ) : Euler’s method
2nd order Taylor series method
yn!! = fy (yn ) · yn! = fy (yn ) · f (yn )
un+1 = un + hf (un ) +
h2
2 fy (un )
· f (un )
ex
y ! = y , y(0) = 1
un+1 = un + hun +
h2
2
*
un = 1 + h +
h2
2
+
un
t = 1 , h = 1/n
h
0.1
0.05
0.025
0.0125
un
2.7140808
2.7171911
2.7180039
2.7182117
y n − un
(yn − un )/h2
0.0042010
0.4201
0.0010907
0.4363
0.0002779
0.4447
0.0000701
0.4488
10
Runge-Kutta methods
y ! = f (y)
k1 = f (un )
k2 = f (un + ahk1 )
un+1 = un + h(bk1 + ck2 ) : 2-stage RK method
choose a , b , c to minimize the local truncation error
yn+1 = yn + h(bf (yn ) + cf (yn + ahf (yn ))) + τn
yn+1 = yn + hyn! +
h2
2
yn!! + O(h3 )
f (yn ) = yn!
f (yn + ahf (yn )) = f (yn ) + fy (yn ) · ahf (yn ) + O(h2 ) = yn! + ahyn!! + O(h2 )
h(byn! + c(yn! + ahyn!! + O(h2 )) + τn = hyn! +
h2
2
yn!! + O(h3 )
h(b + c − 1)yn! + h2 (ac − 12 )yn!! + τn = O(h3 )
b + c − 1 = 0 , ac −
c=
1
2
= 0 ⇒ τn = O(h3 )
1
1
, b=1−
: 1-parameter family of methods
2a
2a
midpoint method : a =
1
2
, b=0 , c=1
k1 = f (un )
k2 = f (un +
h
2 k1 )
un+1 = un + hk2 = un + hf (un +
h
2 f (un ))
modified Euler method : a = 1 , b =
1
2
, c=
1
2
k1 = f (un )
k2 = f (un + hk1 )
un+1 = un +
h
2 (k1
+ k2 ) = un +
h
2 (f (un )
+ f (un + hf (un )))
11
4th order Runge-Kutta
k1 = f (un )
k2 = f (un +
h
2 k1 )
k3 = f (un +
h
2 k2 )
k4 = f (un + hk3 )
un+1 = un +
h
6 (k1
+ 2k2 + 2k3 + k4 ) : 4 stages
ex
y ! = y , y(0) = 1
k1 = un
k2 = un +
h
2 k1
= (1 +
h
2 )un
k3 = un +
h
2 k2
= (1 +
h
2 (1
k4 = un + hk3 = (1 + h(1 +
un+1 = un +
h
6 (1
un+1 = (1 + h +
+ 2(1 +
h2
2
+
h3
6
+
h
2 ))un
+
h
2)
h
2
+
= (1 +
h2
4 ))un
+ 2(1 +
h
2
h
2
+
= (1 + h +
+
h2
4 )
un
2.7182511
2.7182797
2.7182817
h2
2
+
+ (1 + h +
h4
24 )un
t = 1 , h = 1/n
h
0.2
0.1
0.05
h2
4 )un
y n − un
(yn − un )/h4
0.00003069185
0.01918
0.00000208432
0.02084
0.00000013580
0.02173
h3
4 )un
h2
2
+
h3
4 ))un
12
note
Runge-Kutta methods have the form un+1 = un + hF (un , h). Moreover,
(a) F (u, 0) = f (u)
#
#
(b) there exists a constant L,
independent of h, st |F (u, h) − F (v, h)| ≤ L|u
− v|
for all u , v in a given domain and h sufficiently small.
ex : modified Euler method
un+1 = un +
F (u, h) =
h
2 (f (un )
1
2 (f (u)
+ f (un + hf (un )))
+ f (u + hf (u)))
(a) F (u, 0) = f (u)
ok
(b) |F (u, h) − F (v, h)|
=
=
≤
≤
1
2 |f (u)
1
2 |f (u)
1
2 |f (u)
1
2 L|u
=
1
2 L|u
≤
1
2 L|u
+ f (u + hf (u)) − (f (v) + f (v + hf (v)))|
− f (v) + f (u + hf (u)) − f (v + hf (v))|
− f (v)| + 12 |f (u + hf (u)) − f (v + hf (v))|
− v| + 12 L|u + hf (u) − (v + hf (v)|
− v| + 12 L|u − v + h(f (u) − f (v))|
− v| + 12 L|u − v| + 12 Lh|f (u) − f (v)|
≤ L|u − v| + 12 Lh · L|u − v|
= (L + 12 hL2 )|u − v|
#
Then |F (u, h) − F (v, h)| ≤ L|u
− v| for 0 ≤ h ≤ 1, where L# = L + 12 L2 .
hw : midpoint method
thm
1. (a)
2. (b)
⇒
⇒
3. (a) + (b)
τn = O(h2 )
(at least)
the scheme is stable wrt initial data
⇒
the scheme converges
( i.e. lim un = y(t) )
h→0
ok
4. Tues 1/18
13
note
1. (a) ⇒ the difference scheme approximates the correct differential equation
un+1 = un + hF (un , h)
yn+1 = yn + hF (yn , h) + τn
yn+1 − yn
τn
= F (yn , h) +
⇒ yn! = f (yn )
h
h
In this case we say that the difference scheme is consistent with the differential
equation.
2. The scheme is stable wrt initial data if |un − vn | ≤ C|u0 − v0 |, where un , vn
are the numerical solutions starting from u0 , v0 and C is independent of h,
i.e. a small change in the initial data of the difference scheme leads to a small
change in the solution of the difference scheme.
3. The theorem says that consistency + stability ⇒ convergence.
pf
1. un+1 = un + hF (un , h)
yn+1 = yn + hF (yn , h) + τn
yn + hyn! + O(h2 ) = yn + h(F (yn , 0) + O(h)) + τn
ok
2. Let un , vn be the numerical solutions starting from u0 , v0 .
|un+1 − vn1 | = |un + hF (un , h) − (vn + hF (vn , h))|
≤ |un − vn | + h|F (un , h) − F (vn , h)|
⇒
#
≤ (1 + hL)|u
n − vn |
#
#
# n
|un − vn | ≤ (1 + hL)
|u0 − v0 | ≤ enhL |u0 − v0 | = eLt |u0 − v0 |
3. define en = yn − un
ok
en+1 = yn+1 − un+1 = yn + hF (yn , h) + τn − (un + hF (un , h)) = · · ·
#
eLt − 1
|en+1 | ≤ (1 + hL)|en | + τn , . . . |en | ≤ e |e0 | +
· τ , τ = max τn
hL#
#
#
Lt
ok
14
note
1. A method of the form un+1 = un + hF (un , h) is an explicit 1-step method.
This includes Runge-Kutta and Taylor series methods.
2. If τ = O(hp+1 ), then |un − yn | = O(hp ), i.e. the global error is one order lower
than the local truncation error. We say that the method is p-th order accurate.
generalization
1. systems
y1! = f1 (y1 , . . . , yN )
..
.
!
yN = fN (y1 , . . . , yN )





⇒




y1
 . 
!
. 
y = f (y) , y = 
 .  , f =
yN


f1
 . 
 .. 


fN
2. non-autonomous equations
!
y = f (y, t) ,
y1 = y
y2 = t
6
⇒
!
y1
y2
"!
=
y1
y2
"!
!
f (y1 , y2 )
1
"
3. higher order equations
!!
y1 = y
y2 = y !
!
y = f (y, y ) ,
6
⇒
!
=
!
y2
f (y1 , y2 )
"
def : A vector norm ||y|| has the following properties.
1. ||y|| ≥ 0 , ||y|| = 0 ⇔ y = 0
2. ||αy|| = |α| · ||y||
3. ||y + u|| ≤ ||y|| + ||u||
ex
||y||1 =
N
$
i=1
|yi |
,
||y||∞ = max |yi |
,
||y||2 =
note


N
$
i=1
|yi
1/2
|2 
Given a vector norm ||y||, the subordinate matrix norm is ||A|| = max
y+=0
||A|| satisfies properties 1, 2, 3 above and 4, 5 below.
4. ||Ay|| ≤ ||A|| · ||y||
,
5. ||AB|| ≤ ||A|| · ||B||
||Ay||
.
||y||
15
ex
||A||1 = max
j
N
$
i=1
||A||∞ = max
i
|aij | : max column sum
N
$
j=1
|aij | : max row sum
||A||2 = max σi (A) : max singular value , maxi |λ(A)| if A is real symmetric
i
pf : Math 571
note
All previous results for a scalar equation also hold for a system of equations,
with |y| → ||y|| and L = max |fy | → L = max ||fy || , fy = (∂fi /∂yj ).
absolute stability
Consider y ! = λy, where y is a scalar and λ is a complex number (test equation).
The exact solution is y(t) = y(0)eλt , and if real(λ) ≤ 0, then y(t) is bounded as
t → ∞ for all initial data. The analogous property for the numerical method is
absolute stability, i.e. a numerical method with step size h is absolutely stable
if un is bounded as n → ∞ for all initial data, where un is the result of applying
the numerical method to the test equation.
ex : Euler’s method
y ! = λy ⇒ un+1 = un + hλun = (1 + hλ)un ⇒ un = (1 + hλ)n u0
un is bounded as n → ∞ ⇔ |1 + hλ| ≤ 1 ⇔ |hλ − (−1)| ≤ 1
Euler’s method is absolutely stable ⇔ hλ lies in the region of absolute stability
h ! ! plane
!2
!1
16
ex
y ! = Ay : linear system
assume A is diagonalizable : A = XDX −1
D = diag(λ1 , . . . , λN ) , X = [x1 · · · xN ] , Axj = λj xj , j = 1, . . . , N
y(t) = α1 eλ1 t x1 + · · · + αN eλN t xN
un+1 = un + hAun = (I + hA)un
un = (I + hA)n u0 = α1 (1 + hλ1 )n x1 + · · · + αN (1 + hλN )n xN
absolute stability
⇔
|1 + hλj | ≤ 1 , j = 1, . . . , N
Hence to ensure absolute stability for a diagonalizable linear system y ! = Ay, it
is sufficient to ensure absolute stability for the scalar test equation y ! = λy for
each eigenvalue of A.
ex
!
y1
y2
"!
!
−11
9
=
9 −11
"!
y1
y2
"
y(t) = α1 e−20t x1 + α2 e−2t x2
⇒
:
λ1 = −20 , λ2 = −2 : stiff system
fast component + slow component
un = α1 (1 − 20h)n x1 + α1 (1 − 2h)n x2
absolute stability
⇔
|1 − 20h| ≤ 1 , |1 − 2h| ≤ 1
⇔
−1 ≤ 1 − 20h ≤ 1 , −1 ≤ 1 − 2h ≤ 1
⇔
−1 ≤ 1 − 20h , −1 ≤ 1 − 2h
⇔
h ≤ 0.1 , h ≤ 1
Hence we must choose h ≤ 0.1 to ensure that Euler’s method is absolutely stable.
note
The fast component of the exact solution decays rapidly, and after some time,
essentially only the slow component is present, i.e. y(t) ∼ α2 e−2t x2 . The slow
component by itself requires only h ≤ 1 for absolute stability, but we must choose
h ≤ 0.1 to ensure absolute stability of Euler’s method applied to the system.
17
ex :
!
y1
y2
"!
!
−11
9
=
9 −11
"!
y1
y2
"
,
!
y1
y2
"
0
=
2
0
3
3
2
2
1
1
0
0
!1
!1
!2
!2
!3
!3
0.5
1
1.5
2
3
3
2
2
1
1
0
0
!1
!1
!2
!2
!3
!3
0.5
1
1.5
2
0
0
forward Euler , h = 0.09 , AS
3
2
2
1
1
0
0
!1
!1
!2
!2
0
0.5
1
1.5
0.5
1
1.5
2
0.5
1
1.5
2
backward Euler , h = 0.09 , AS
3
!3
, solid line = y1 (t)
backward Euler , h = 0.1 , AS
forward Euler , h = 0.1 , AS
0
"
backward Euler , h = 0.1025 , AS
forward Euler , h = 0.1025 , not AS
0
!
2
!3
0
0.5
1
1.5
2
5. Thurs 1/20
18
backward Euler method
un+1 − un
= f (un+1 ) ⇒ un+1 = un + hf (un+1 ) : implicit
h
note : forward Euler is explicit
claim : τn = O(h2 )
pf : hw
test equation : y ! = λy
un+1 = un + hλun+1
absolute stability ⇔
⇒
%
%
%
%
%1
(1 − hλ)un+1 = un
%
1 %%
% ≤ 1
− hλ %
⇔
⇒
un+1 =
1
un
1 − hλ
|1 − hλ| ≥ 1
h ! ! plane
1
note
1. A scheme is called A-stable if the region of absolute stability contains the left
half-plane. Backward Euler is A-stable and forward Euler is not A-stable. For
an A-stable scheme, the stepsize h is not restricted by absolute stability and it
can be chosen entirely on the basis of accuracy requirements.
2. un+1 = un + hf (un+1 ) can be solved by iteration.
(0)
(m+1)
un+1 = un + hf (un ) , un+1
(m)
= un + hf (un+1 ) : predictor-corrector
Thus implicit schemes require more work per timestep than explicit schemes.
3. un+1 = un + h2 (f (un ) + f (un+1 )) : trapezoid method , implicit
claim : τn = O(h3 ) , A-stable
pf : hw
19
region of absolute stability for 4th order Runge-Kutta
k1 = f (un ) , k2 = f (un + h2 k1 ) , k3 = f (un + h2 k2 ) , k4 = f (un + hk3 )
un+1 = uu + h6 (k1 + 2k2 + 2k3 + k4 )
y ! = λy
⇒
un+1


h2 λ2 h3 λ3 h4 λ4 

= 1 + hλ +
+
+
un
2
6
24
4
3
2
1
0
!1
!2
!3
!4
!5
!4
!3
!2
!1
0
1
The region of absolute stability for RK4 :
7
1. is larger than the region for forward Euler,
2. contains an interval on the imaginary axis.
2
3
20
multistep methods
Adams-Bashforth
un+1 = un + h(β0 f (un ) + β1 f (un−1 ) + · · · + βk f (un−k ))
: (k + 1)-step method , explicit
idea
!
y = f (y)
⇒
yn+1 = yn +
8 t
n+1
tn
f (y(t)) dt
approximate f (y(t)) by an interpolating polynomial p(t)
p(tn ) = f (un )
p(tn−1 ) = f (un−1 )
..
.
p(tn−k ) = f (un−k )
def
∇fn = fn − fn−1 : backward difference
∇2 fn = ∇(∇fn ) = ∇(fn − fn−1 ) = ∇fn − ∇fn−1 = fn − fn−1 − (fn−1 − fn−2 )
= fn − 2fn−1 + fn−2
..
.
Ifn = fn , S− fn = fn−1 ⇒ ∇ = I − S−
∇fn
∇2 fn
p(t) = fn + (t − tn )
+ (t − tn )(t − tn−1 ) 2
h
2h
∇k fn
+ · · · + (t − tn ) · · · (t − tn−k+1 )
k! hk
&
'(
)
k terms
note
p(t) is a polynomial of degree k , it resembles a Taylor polynomial
∇i fn
p(t) =
(t − tn ) · · · (t − tn−i+1 )
i! hi
i=0
k
$
6. Tues 1/25
21
claim
1. p(tn−j ) = fn−j
,
j = 0, . . . , k : k + 1 interpolation points
(t − tn ) · · · (t − tn−k ) dk+1
2. f (y(t)) − p(t) =
f (y(α)) for some α = α(t)
(k + 1)!
dtk+1
note
p(t) interpolates f (un−j ) in 1 and f (yn−j ) in 2.
pf 1.
p(tn ) = fn
p(tn−1 ) = fn + (tn−1 − tn )
= fn + (−h)
∇fn
h
∇fn
= fn − ∇fn = fn − (fn − fn−1 ) = fn−1
h
p(tn−2 ) = fn + (tn−2 − tn )
∇fn
∇2 fn
+ (tn−2 − tn )(tn−2 − tn−1 ) 2
h
2h
∇fn
∇2 fn
= fn + (−2h)
+ (−2h)(−h) 2
h
2h
= fn − 2∇fn + ∇2 fn = fn − 2(fn − fn−1 ) + (fn − 2fn−1 + fn−2 ) = fn−2
∇i fn
p(tn−j ) = (tn−j − tn )(tn−j − tn−1 ) · · · (tn−j − tn−i+1 )
i! hi
i=0
j
$
(check i = 0)
∇i fn
= (−jh)(−(j − 1)h) · · · (−(j − i + 1)h)
i! hi
i=0
j
$
∇i fn
= (−1)i · hi · j(j − 1) · · · (j − i + 1)
=
i
i!
h
i=0
j
$
recall : binomial expansion
j
$
 
j
 aj−i bi
(a + b)j =
i=0 i
⇒ p(tn−j ) =
j
$
i=0
 
,
 
(−1)i 
i=0
j i
∇ fn
i
 
j
j!
=
: binomial coefficient
i
i!(j − i)!
 
j
 I j−i (−∇)i f
n
i
j
$
= (I − ∇)j fn = S−j fn = fn−j
ok
22
recall
8 t
n+1
yn+1 = yn +
un+1 = un +
t − tn
h
s =
tn
8 t
f (y(t)) dt
n+1
tn
p(t) dt
⇒ t − tn = sh
t − tn−1 = t − tn + tn − tn−1 = sh + h = (s + 1)h
..
.
t − tn−k+1 = (s + k − 1)h
p(t) = fn + s∇fn +
8 t
n+1
tn
p(t) dt =
8 1
0
s(s + 1) 2
s(s + 1) · · · (s + k − 1) k
∇ fn + · · · +
∇ fn
2
k!
9
p(t(s))hds = h γ0 fn + γ1 ∇fn + γ2 ∇2 fn + · · · + γk ∇k fn
γ0 =
8 1
ds = 1
γ1 =
8 1
s ds =
γ2 =
8 1
s(s + 1)
ds =
2
8 1
s(s + 1) · · · (s + k − 1)
ds , k ≥ 1
k!
..
.
γk =
0
0
0
0
:
1
2
5
12
k = 0 : 1-step AB
un+1 = un + hγ0 fn = un + hfn : Euler’s method
k = 1 : 2-step AB
9
:
un+1 = un +h (γ0 fn + γ1 ∇fn ) = un +h fn + 12 (fn − fn−1 ) = un + h2 (3fn −fn−1 )
k = 2 : 3-step AB
9
:
un+1 = un + h γ0 fn + γ1 ∇fn + γ2 ∇2 fn = un +
h
12 (23fn
− 16fn−1 + 5fn−2 )
23
note
1. To get started, we set u0 = y0 , but we also need to compute u1 , u2 , . . . , uk .
2. To compute un+1 from un , only 1 function evaluation fn = f (un ) is needed.
3. Changing the step size h requires interpolation or variable coefficient formulas.
pf 2.
(t − tn ) · · · (t − tn−k ) dk+1
goal : f (y(t)) = p(t) +
f (y(α))
(k + 1)!
dtk+1
assume t += tn−j , j = 0, . . . , k
g(x) = f (y(x)) − p(x) +
(x − tn ) · · · (x − tn−k )
( p(t) − f (y(t)) )
(t − tn ) · · · (t − tn−k )
g(tn ) = 0 , g(tn−1 ) = 0 , . . . , g(tn−k ) = 0 , g(t) = 0
⇒
g(x) has k + 2 distinct roots
⇒
g ! (x) has k + 1 distinct roots
..
.
⇒
g (k+1) (x) has 1 root , say g (k+1) (α) = 0
then 0 = g (k+1) (α) =
( MVT : Math 451 )
dk+1
(k + 1)!( p(t) − f (y(t)) )
f
(y(α))
+
dtk+1
(t − tn ) · · · (t − tn−k )
local truncation error : (k + 1)-step AB
yn+1 = yn +
un+1 = un +
yn+1 = yn +
; tn+1
tn
; tn+1
tn
; tn+1
tn
τn = yn+1 − yn −
=
8 t
n+1
=
8 1
tn
0
f (y(t)) dt
p(t) dt
p(t) dt + τn
; tn+1
tn
p(t) dt =
; tn+1
tn ( f (y(t)
− p(t) ) dt
(t − tn ) · · · (t − tn−k ) dk+1
f (y(α)) dt
(k + 1)!
dtk+1
s(s + 1) · · · (s + k)hk+1 (k+2)
# hk+2
y
(α) h ds = γk+1 y (k+2) (α)
(k + 1)!
ok
7. Thurs 1/27
24
Adams-Moulton
∗
un+1 = un + h(β−1
f (un+1 ) + β0∗ f (un ) + · · · + βk∗ f (un−k ))
: (k + 1)-step method , implicit
p∗ (t) interpolates fn+1 , fn , . . . , fn−k
∗
p (t) = fn+1
:
k + 2 values
∇fn+1
∇2 fn+1
+ (t − tn+1 )
+ (t − tn+1 )(t − tn )
h
2h2
+ · · · + (t − tn+1 ) · · · (t − tn+1−k )
&
'(
k + 1 terms
)
∇k+1 fn+1
(k + 1)! hk+1
t − tn = sh ⇒ t − tn+1 = t − tn + tn − tn+1 = sh − h = (s − 1)h
p∗ (t) = fn+1 + (s − 1)∇fn+1 +
(s − 1)s 2
∇ fn+1
2
+ ··· +
un+1 = un +
8 t
n+1 ∗
p (t) dt
tn
=
(s − 1) · · · (s + k − 1) k+1
∇ fn+1
(k + 1)!
8 1
p∗ (t(s))hds
0
9
∗
un+1 = un + h γ−1
fn+1 + γ0∗ ∇fn+1 + · · · + γk∗ ∇k+1 fn+1
γk∗ =
∗
γ−1
8 1
(s − 1)s · · · (s + k − 1)
ds , k ≥ 0
(k + 1)!
0
=
8 1
0
γ0∗ =
8 1
γ1∗ =
8 1
0
0
ds = 1
(s − 1) ds = − 12
(s − 1)s
1
ds = − 12
2
k = −1
∗
un+1 = un + hγ−1
fn+1 = un + hfn+1 : backward Euler
:
25
k = 0 : 1-step AM
9
:
∗
un+1 = un + h (γ−1
fn+1 + γ0∗ ∇fn+1 ) = un + h fn+1 − 12 (fn+1 − fn )
un+1 = un + h2 (fn+1 + fn ) : trapezoid method
k = 1 : 2-step AM
∗
un+1 = un + h (γ−1
fn+1 + γ0∗ ∇fn+1 + γ1∗ ∇fn+1 )
un+1 = un +
h
12 (5fn+1
+ 8fn − fn−1 )
summary
AB
9
un+1 = un + h γ0 fn + γ1 ∇fn + · · · + γk ∇k fn
γk =
8 1
0
:
: k + 1-step
s(s + 1) · · · (s + k − 1)
ds , k ≥ 1 , γ0 = 1
k!
τn = γk+1 y (k+2) (t) hk+2 + O(hk+3 )
AM
9
∗
un+1 = un + h γ−1
fn+1 + γ0∗ ∇fn+1 + · · · + γk∗ ∇k+1 fn+1
γk∗ =
8 1
0
(s − 1)s · · · (s + k − 1)
∗
ds , k ≥ 0 , γ−1
= 1
(k + 1)!
∗
τn = γk+1
y (k+3) (t) hk+3 + O(hk+4 )
note
k-step AB has global order of accuracy k
k-step AM · · · · · · · · · · · · ” · · · · · · · · · · · · k + 1
:
: k + 1-step
26
region of absolute stability
1
AB
hλ-plane
0.8
0.6
0.4
0.2
6
0
5
4
!0.2
!0.4
3
!0.6
!0.8
2
1
!1
!2
!1.5
!1
!0.5
0
0.5
1
5
4
AM
hλ-plane
3
2
1
0
3
4
5
6
!1
!2
!3
!4
!5
!8
!6
!4
!2
0
2
4
The method of order k is absolutely stable inside the contour. Note the difference
in scale; the regions are much smaller for AB than for AM. (1-step and 2-step
AM are not shown because they are A-stable)
27
general multistep methods
α0 un + α1 un−1 + · · · + αk un−k + h(β0 f (un ) + β1 f (un−1 ) + · · · + βk f (un−k )) = 0
: k-step method
k
$
(αi un−i + hβi f (un−i )) = 0
i=0
predictor : β0 = 0
α0 u(0)
n
=−
k
$
(αi un−i + hβi f (un−i ))
i=1
corrector
α0 u(m+1)
n
=
−hβ0 f (u(m)
n )
−
k
$
(αi un−i + hβi f (un−i ))
i=1
local truncation error : (general case)
k
$
(αi yn−i + hβi f (yn−i )) = τn = O(hr+1 )
i=0
yn−i
y (j) (t)
= y(t − ih) =
(−ih)j + O(hr+2 )
j!
j=0
τn =
r+1
$
k
$
i=0
!
(αi yn−i + hβi yn−i
)


r+1
$ y (j+1) (t)
y (j) (t)
j

=
αi
(−ih) + hβi
(−ih)j  + O(hr+2 )
j!
j!
i=0
j=0
j=0
k
$

r+1
$

r+1
$ y (j) (t)
y (j) (t)
j

=
αi
(−ih) + hβi
(−ih)(j−1)  + O(hr+2 )
j!
i=0
j=0
j=1 (j − 1)!
k
$
τn =
r+1
$
r+1
$
Cj y (j) (t)hj + O(hr+2 )
j=0
C0 =
k
$
i=0
αi
,


(−i)j αi (−i)(j−1) βi 

Cj =
+
j!
(j − 1)!
i=0
k
$
If C0 = · · · = Cr = 0 , Cr+1 += 0 , then τn = Cr+1 y (r+1) (t)hr+1 + O(hr+2 ) and
we have a k-step method of order r.
8. Tues 2/1
28
ex : leap-frog method
y ! = f (y)
un+1 − un−1
= f (un )
2h
un+1 − un−1 − 2hf (un ) = 0 : explicit , 2-step , needs u0 , u1 to start
α0 = 1 , α1 = 0 , α2 = −1 , β0 = 0 , β1 = −2 , β2 = 0
C0 =
k
$
αi = α0 + α1 + α2 = 0
k
$
(−iαi + βi ) = −α1 − 2α2 + β0 + β1 + β2 = 0
i=0
C1 =
i=0


i2 αi
α1

C2 =
− iβi  =
+ 2α2 − β1 − 2β2 = 0
2
2
i=0
k
$


−i3 αi i2 βi  −α1
4
1
1

C3 =
+
=
− α2 + β1 + 2β2 =
6
2
6
3
2
3
i=0
k
$
1 (3)
y (t)h3 + O(h4 ) ⇒ leap-frog has order 2
3
question : What is the maximum order of a k-step multistep scheme?
τn =
α0 , . . . , αk , β0 , . . . , βk : 2k + 2 unknown coefficients
⇒ 2k + 1 degrees of freedom
⇒ C0 = · · · = C2k = 0 , C2k+1 += 0
⇒ τn = O(h2k+1 ) is optimal
⇒ the maximum order of a k-step multistep scheme is 2k
ex
k = 1 : trapezoid method has order 2
k = 2 : 2-step AB has order 2 , 2-step AM has order 3
In fact, there exists a 2-step method of order 4. (more later)
note : RK4 is a 1-step scheme of order 4, but there is no contradiction because
RK4 is not a multistep scheme of the form considered here
29
characteristic polynomials of a multistep scheme
α0 un + α1 un−1 + · · · + αk un−k + h(β0 f (un ) + β1 f (un−1 ) + · · · + βk f (un−k )) = 0
k
def : ρ(ζ) = α0 ζ + α1 ζ
k−1
+ · · · + αk =
σ(ζ) = β0 ζ k + β1 ζ k−1 + · · · + βk =
k
$
αi ζ k−i
k
$
βi ζ k−i
i=0
i=0
note
ρ(1) =
k
$
αi = C0
k
$
βi
k
$
(k − i)αi = k
i=0
σ(1) =
i=0
ρ! (1) =
i=0
k
$
i=0
αi +
k
$
(−iαi + βi ) −
i=0
k
$
i=0
βi = kC0 + C1 − σ(1)
recall : a method is consistent ⇔ τn = O(hr+1 ) for some r ≥ 1
⇔ C0 = C1 = 0
⇔ ρ(1) = 0 , ρ! (1) + σ(1) = 0
ex : leap-frog
un − un−2 − 2hf (un−1 ) = 0 ⇒ ρ(ζ) = ζ 2 − 1 , σ(ζ) = −2ζ
ρ(1) = 0 , ρ! (1) + σ(1) = 0 ⇒
leap-frog is consistent
questions
When is a general multistep scheme stable? convergent? absolutely stable?
special case : y ! = 0
k
$
αi un−i = 0 : linear , constant coefficient , homogeneous difference equation
i=0
α0 un + α1 un−1 + · · · + αk un−k = 0 (∗)
given u0 , u1 , . . . , uk−1 , use (∗) to solve for uk , uk+1 , . . .
30
note
If we set un = ζ n , then
k
$
αi un−i =
i=0
k
$
αi ζ n−i = ζ n−k
i=0
k
$
αi ζ k−i = ζ n−k ρ(ζ).
i=0
Hence if ρ(ζ) = 0, then un = ζ n is a solution of (∗).
thm
If ρ(ζ) has j distinct roots ζ1 , . . . , ζj with multiplicities m1 , . . . , mj , then the
general solution of the difference equation (∗) has the form
un = (a10 + a11 n + a12 n2 + · · · + a1,m1 −1 nm1 −1 )ζ1n
..
.
+ (aj0 + aj1 n + aj2 n2 + · · · + aj,mj −1 nmj −1 )ζjn ,
where a10 , . . . , aj,mj −1 are determined by the initial data u0 , . . . , uk−1 .
def
ζ1 , . . . , ζj are the characteristic roots of the difference equation (∗)
ex
1. un + 4un−1 − 5un−2 = 0
ρ(ζ) = ζ 2 + 4ζ − 5 = (ζ − 1)(ζ + 5) ⇒ ζ1 = 1 , ζ2 = −5 , m1 = m2 = 1
un = a1 ζ1n + a2 ζ2n = a1 + a2 (−5)n
check . . .
2. un − 2un−1 + un−2 = 0
ρ(ζ) = ζ 2 − 2ζ + 1 = (ζ − 1)2 ⇒ ζ1 = 1 , m1 = 2
un = (a1 + a2 n)ζ1n = a1 + a2 n
check . . .
note
There is an analogy between difference equations and differential equations.
α0 y (k) + α1 y (k−1) + · · · + αk y = 0 , y(t) = eλt ⇒ ρ(λ) = 0
1. y !! +4y ! −5y = 0 ⇒ λ1 = 1 , λ2 = −5 , m1 = m2 = 1 ⇒ y(t) = a1 et +a2 e−5t
2. y !! − 2y ! + y = 0 ⇒ λ1 = 1 , m1 = 2 ⇒ y(t) = (a1 + a2 t)et
9. Thurs 2/3
31
pf (sketch)
α0 un + α1 un−1 + · · · + αk un−k = 0
α1
αk
α1
αk
un = − un−1 − · · · −
un−k ⇒ uk = − uk−1 − · · · −
u0
α0
α0
α0
α0


 α1

αk  
uk
uk−1
−
·
·
·
·
·
·
−


 α0


α0 









u

 k−1 






 .

 ..









=
u1













U1 = AU0
,
1
0
...
...
1
0
⇒
Un = AUn−1




u

  k−2 






 . 
  .. 








u0
Un = An U0
A = XJX −1 : Jordan form
,
(−1)k
det(A − λI) =
ρ(λ) ⇒
α0
e-values of A are characteristic roots of (∗)
J=




J1
0
...
0
Jj




note
, Ji =







ζi
An = XJ n X −1
1
...
...
...
0
0
1
ζi







: mi × mi
ok
the difference scheme (∗) is absolutely stable
⇔ every solution un of (∗) is bounded as n → ∞
⇔ ρ(ζ) satisfies the root condition :




7
1. |ζi | ≤ 1 for i = 1, . . . , j
2. if |ζi | = 1 , then mi = 1
0 if |ζ| < 1
pf : n→∞
lim n |ζ| =  1 if |ζ| = 1 , p = 0


∞ if |ζ| > 1 or |ζ| = 1 , p ≥ 1
p
ex
n
ρ(ζ)
root condition?
un
ζ + 4ζ − 5
no
a1 + a2 (−5)n
ζ 2 − 2ζ + 1
no
a1 + a2 n
2
ζ −1
yes
a1 + a2 (−1)n
2
32
test equation : y ! = λy
k
$
i=0
(αi un−i + hβi f (un−i )) = 0 ⇒
k
$
(αi + hλβi )un−i = 0
i=0
(α0 + hλβ0 )un + (α1 + hλβ1 )un−1 + · · · + (αk + hλβk )un−k = 0 (∗∗)
note
If we set un = ζ n , then
k
$
(αi + hλβi )un−i = · · · = ζ n−k (ρ(ζ) + hλσ(ζ)).
i=0
Hence if ρ(ζ) + hλσ(ζ) = 0, then un = ζ n is a solution of (∗∗).
claim
1. If ρ(1) = 0, ρ! (1) += 0, then ρ(ζ) + hλσ(ζ) has a root ζ1 (h) st lim ζ1 (h) = 1.
h→0
r+1
2. If in addition τn = O(h
hλ
), then ζ1 (h) = e
r+1
+ O(h
).
note
1. ρ(1) = 0 , ρ! (1) += 0 ⇒ ρ(ζ) has a simple root at ζ = 1
This holds if the scheme is consistent and ρ(ζ) satisfies the root condition.
2. ζ1 (h) is called the principal root of the difference scheme (∗∗); the other roots
of ρ(ζ) + hλσ(ζ) are denoted ζ2 (h), . . . and are called the extraneous roots. For
fixed t = nh, (∗∗) has a solution of the form un = ζ1 (h)n = (ehλ + O(hr+1 ))n =
eλt + O(hr ). This justifies calling ζ1 (h) the principal root.
ex : 2-step AB
y ! = λy
,
un − un−1 − 12 hλ(3un−1 − un−2 ) = 0
ρ(ζ) = ζ 2 − ζ
ρ(1) = 0 , ρ! (1) += 0 , τn = O(h3 )
⇒
ρ(ζ) + hλσ(ζ) = ζ 2 − ζ − 12 hλ(3ζ − 1)
ζ1 (h) =
1
2
+ 34 hλ +
1
2
ζ2 (h) =
1
2
+ 34 hλ −
1
2
9
1 + hλ + 94 h2 λ2
9
1 + hλ + 94 h2 λ2
:1/2
:1/2
ζ1 (h)n = (ehλ + O(h3 ))n = eλt + O(h2 )
ζ2 (h)n = O(hn )
un = a1 ζ1 (h)n + a2 ζ2 (h)n
= ehλ + O(h3 )
= O(h)
⇒
claim applies
10. Tues 2/8
33
Consider y(0) = 1 , y(t) = eλt . Take u0 = 1 , u1 = ehλ .
!
1
ζ1
1
ζ2
"!
a1
a2
"
!
1
= hλ
e
"
!
⇒
a1
a2
"
1
=
ζ1 − ζ2
! hλ
e − ζ2
ζ1 − ehλ
ehλ − ζ2
ehλ − ζ1 + ζ1 − ζ2
a1 =
=
= 1 + O(h3 ) ,
ζ1 − ζ2
ζ1 − ζ2
ζ1 − ehλ
a2 =
= O(h3 )
ζ1 − ζ2
9
:
9
"
ζ1 − ζ2 = O(1)
:
un = 1 + O(h3 ) · eλt + O(h2 ) + O(h3 ) · O(hn )
⇒ lim un = eλt , i.e. the 2-step AB scheme converges for fixed t = nh
h→0
ex : leap-frog
y ! = λy
,
un − un−2 − 2hλun−1 = 0
ρ(ζ) = ζ 2 − 1
ρ(1) = 0 , ρ! (1) += 0 , τn = O(h3 )
⇒
⇒
ρ(ζ) + hλσ(ζ) = ζ 2 − 1 − 2hλζ
claim applies
ζ1 (h) = hλ + (1 + h2 λ2 )1/2 = ehλ + O(h3 )
ζ2 (h) = hλ − (1 + h2 λ2 )1/2 = −e−hλ + O(h3 )
ζ1 (h)n = (ehλ + O(h3 ))n = eλt + O(h2 )
ζ2 (h)n = ( − e−hλ + O(h3 ))n = (−1)n e−λt + O(h2 )
9
:
9
:
9
:
un = 1 + O(h3 ) · eλt + O(h2 ) + O(h3 ) · (−1)n e−λt + O(h2 )
⇒ lim un = eλt , i.e. the leap-frog scheme converges for fixed t = nh
h→0
note : if h is fixed and n → ∞, then O(h3 )(−1)n e−λt →
7
0
if real(λ) > 0
±∞ if real(λ) < 0
This is an example of weak instability, which arises when ρ(ζ) has two distinct
roots on the unit circle. In this case, the roots of ρ(ζ) + hλσ(ζ) may lie inside or
outside the unit circle. For the leap-frog method with real(λ) > 0, the extraneous
root ζ2 (h) lies inside the unit circle and is harmless, but with real(λ) < 0, ζ2 (h)
lies outside the unit circle and a2 ζ2 (h)n eventually dominates a1 ζ1 (h)n , even
though a2 = O(h3 ).
34
weak instability of the leap-frog method
y ! = λy , y(0) = 1 ,
un+1 = un−1 + 2hλun , u0 = 1 , u1 = ehλ
! = 1 , h = 0.5
! = !1 , h = 0.5
150
1
100
0.5
50
0
0
0
1
2
3
4
5
!0.5
0
1
! = 1 , h = 0.25
1
100
0.5
50
0
0
1
2
3
4
5
!0.5
0
1
! = 1 , h = 0.125
1
100
0.5
50
0
0
1
2
3
4
5
2
3
4
5
4
5
! = !1 , h = 0.125
150
0
3
! = !1 , h = 0.25
150
0
2
4
5
!0.5
0
1
2
3
35
absolute stability of leap-frog method
y ! = λy , un+1 = un−1 + 2hλun
what do we know already?
ρ(ζ) = ζ 2 − 1 = 0 ⇒ ζ = ±1 : root condition is satisfied
ρ(ζ) + hλσ(ζ) = ζ 2 − 1 − 2hλζ = (ζ − ζ1 )(ζ − ζ2 )
ζ1 = ehλ + O(h3 ) , ζ2 = −e−hλ + O(h3 )
If we ignore the O(h3 ) terms, then (a) |ζ1 | = |ζ2 | = 1 ⇔ λ is purely imaginary,
(b) if |ζ1 | < 1, then |ζ2 | > 1, and if |ζ1 | > 1, then |ζ2 | < 1. Now we will make
this more precise.
!
"
2
9
:1/2
ζ
−
1
1
1
ζ = hλ ± 1 + h2 λ2
, hλ =
=
ζ−
: conformal mapping
2ζ
2
ζ
ζ-plane
hλ-plane
i
1
−i
1. the branch points hλ = ±i are fixed by the mapping, and every other point
in the hλ-plane corresponds to 2 points in the ζ-plane, e.g. hλ = 0 ⇔ ζ = ±1
9
2. hλ = iy , |y| ≤ 1 ⇒ ζ = iy ± 1 − y 2
:1/2
: unit circle in ζ-plane
:
1 9 iθ
e − e−iθ = i sin θ : imaginary interval in hλ-plane
2
3. For any value of hλ away from the interval, the corresponding values of ζ are
away from the circle. In fact ζ1 ζ2 = −1, so one root lies inside the circle and the
other root lies outside. Hence the region of absolute stability of the leap-frog
method is the interval in the hλ-plane between −i and i, excluding ±i.
ζ = eiθ ⇒ hλ =
11. Thurs 2/10
36
pf (claim)
1. set F (ζ, h) = ρ(ζ) + hλσ(ζ) = 0 , F (1, 0) = ρ(1) = 0 , Fζ (1, 0) = ρ! (1) += 0
implicit function thm ⇒ there exists ζ1 (h) st F (ζ1 (h), h) = 0 , lim ζ1 (h) = 1
h→0
ok
λt
λnh
2. set y(t) = e = e
τn =
k
$
(αi + hλβi )yn−i =
i=0
= eλ(n−k)h
k
$
(αi + hλβi )eλ(n−i)h
i=0
k
$
9
:
(αi + hλβi )eλ(k−i)h = eλ(n−k)h ρ(ehλ ) + hλσ(ehλ )
i=0
ρ(ehλ ) + hλσ(ehλ ) = O(hr+1 )
⇒
set ζ1 (h) = ehλ + " , lim ζ1 (h) = 1 ⇒
lim " = 0
h→0
h→0
0 = ρ(ζ1 (h)) + hλσ(ζ1 (h)) = ρ(ehλ + ") + hλσ(ehλ + ")
= ρ(ehλ ) + ρ! (ehλ )" + O("2 ) + hλσ(ehλ ) + O(h")
= O(hr+1 ) + O(") + O("2 ) + O(h")
⇒
" = O(hr+1 )
ok
convergence theory
y ! = f (y) , y(0) = y0 : well-posed IVP
k
$
(αi un−i + hβi f (un−i )) = 0 , u0 , u1 , . . . , uk−1 : given
i=0
def
1. A multistep method is convergent if un → y(t) as n → ∞, under the assumptions that t = nh is fixed and u0 , u1 , . . . , uk−1 → y0 .
2. A multistep method is stable wrt initial data if there exists a constant C st
for all n ≥ k, |un − vn | ≤ C · max{|u0 − v0 |, . . . , |uk−1 − vk−1 |}, where un , vn are
two solutions of the difference scheme, t = nh is fixed, and h is sufficiently small.
The constant C may depend on t, but must be independent of n.
thm
1. stability ⇔ root condition for ρ(ζ)
2. if the scheme is consistent, then stability ⇔ convergence
37
ex
un + 4un−1 − 5un−2 − h(4f (un−1 + 2f (un−2 )) = 0 : 2-step , explicit
hw : τn = O(h4 )
⇒
the scheme is consistent
ρ(ζ) = ζ 2 + 4ζ − 5 = (ζ − 1)(ζ + 5) : root condition fails
⇒
the scheme is unstable
⇒
the scheme is not convergent
consider y ! = −y , y(0) = 1 ⇒ y(t) = e−t
,
u0 = 1 , u1 = e−hλ
h=0.2
h=0.05
10
10
5
5
0
0
!5
!5
!10
0
0.2
0.4
0.6
0.8
1
!10
0
0.2
h=0.1
10
5
5
0
0
!5
!5
0
0.2
0.4
0.6
0.6
0.8
1
0.8
1
h=0.025
10
!10
0.4
0.8
1
!10
0
0.2
0.4
0.6
For any step size h, the numerical solution is accurate for some time, but eventually the growing oscillations of the extraneous solution destroy the accuracy;
this occurs at an earlier
time as: h is reduced.
This can be seen
in the expression
9
9
:
4
−t
3
4
n 3t/5
un = (1 + O(h )) · e + O(h ) + O(h ) · (−5) e
+ O(h) (hw). This strong
instability is in contrast to weak instability (e.g. leap-frog) where the growing
oscillations of the extraneous solution appear at a later time as h is reduced.
38
recall : A k-step scheme has order ≤ 2k.
thm (Dahlquist)
7
≤ k + 1 if k is odd,
≤ k + 2 if k is even.
If the scheme has order k + 2, then the roots of ρ(ζ) all lie on the unit circle
(and so the method is weakly unstable).
A stable k-step scheme has order
note
k = 1 : the trapezoid method is a stable 1-step scheme of order 2
k = 2 : Milne’s method is a stable 2-step scheme of order 4 : hw
recall : A multistep method is A-stable if the region of absolute stability
contains the left half-plane (e.g. backward Euler, trapezoid method).
thm (Dahlquist)
An A-stable multistep method has order ≤ 2. Among all A-stable schemes of
1
order 2, the trapezoid method has the smallest value of C3 (= 12
).
def
A multistep method is A(α)-stable for 0 ≤ α ≤ π/2 if the region of absolute
stability contains the wedge | arg(hλ) − π| ≤ α.
plane
hλ
α
thm (Widlund)
For any k ≤ 4 and 0 ≤ α ≤ π/2, there exists a k-step method of order k which
is A(α)-stable.
12. Tues 2/15
39
def
A multistep method is stiffly stable if the region of absolute stability contains a
domain of the type shown below.
hλ
plane
ex : BDF methods (backward differentiation formulas)
p(t) = un + (t − tn )
∇un
∇2 un
+ (t − tn )(t − tn−1 )
≈ y(t)
h
2h2
∇un
∇ 2 un
p (t) =
+ (2t − tn − tn−1 )
h
2h2
!
p! (tn ) =
∇un ∇2 un
+
≈ y ! (tn )
h
2h
∇un + 12 ∇2 un = hf (un ) : 2-step BDF , implicit
3
2 un
− 2un−1 + 12 un−2 − hf (un ) = 0
τn = O(h3 ) : 2nd order accurate
ρ(ζ) = 32 ζ 2 − 2ζ +
1
2
= 12 (ζ − 1)(3ζ − 1)
⇒
ζ1 = 1 , ζ2 =
thm (Gear)
The k-step BDF method is stiffly stable for k ≤ 6.
1
3
: stable
40
region of absolute stability for BDF methods
8
hλ-plane
6
4
2
1
0
2
3
4
!2
!4
!6
6
!8
!6
5
!4
!2
0
2
4
6
8
10
12
For k = 1 : 6, the k-step BDF method is absolutely stable outside the contour.
miscellaneous
1. implicit schemes : F (un ) = 0 , fixed-point iteration , Newton’s method
2. other methods for y ! = f (y) : implicit RK , Radau , . . .
3. software : www.netlib.org , Matlab : ode23 , ode45 , . . .
5. adaptive error control : variable timstep/order
6. mechanical systems : M X !! + CX ! + KX = F : Newmark , HHT , . . .
7. Hamiltonian systems :
7 !
p = Hq
q ! = −Hp
, geometric/symplectic methods
8. text : BVP for ODEs , DAE (differential-algebraic equations)
41
2. IBVP for PDEs
heat equation
(chap. 2)
temperature : v(x, t)
,
heat flux : κ∇v(x, t)
conservation of energy
8
d 8
v dx =
κ∇v · n dS ⇒
∂D
dt D
model problem
8
D
vt dx =
8
D
∇ · κ∇v dx ⇒ vt = ∇ · κ∇v
vt = vxx , v(x, 0) = f (x) : find v(x, t) for t > 0
1. −∞ < x < ∞ : free-space
2. 0 ≤ x ≤ 1 , Dirichlet : v(0, t) = v(1, t) = 0
Neumann : vx (0, t) = vx (1, t) = 0
periodic : v(0, t) = v(1, t) , vx (0, t) = vx (1, t)
We assume that the problem is well-posed. (Math 454, 556, 656)
finite-difference scheme
h = ∆x , xj = jh , j = 0, ±1, ±2, . . .
k = ∆t , tn = nk , n = 0, 1, 2, . . .
v(xj , tn ) ≈
unj
,
un+1
− unj
unj+1 − 2unj + unj−1
j
=
k
h2
stencil
n+1
n
j−1
j
j+1
42
def : λ =
k
h2
h=k=
⇒
1
4
un+1
= unj + λ(unj+1 − 2unj + unj−1 )
j
⇒ λ=4
h=
1
4
, k=
1
8
⇒ λ=2
note
D+ unj
unj+1 − unj
unj − unj−1
n
=
, D − uj =
h
h
⇒ un+1
= unj + kD+ D− unj
j
def
||un ||∞ = max |unj |
j
The scheme is stable if ||un ||∞ ≤ ||u0 ||∞ for all n ≥ 1 and all u0 . This property
is also called the maximum principle.
claim : the scheme is stable ⇔ λ ≤
1
2
pf
un+1
= λunj+1 + (1 − 2λ)unj + λunj−1
j
⇐)
|un+1
| ≤ λ|unj+1 | + (1 − 2λ)|unj | + λ|unj−1 |
j
≤ λ||un ||∞ + (1 − 2λ)||un ||∞ + λ||un ||∞ = ||un ||∞
⇒
⇒)
||un+1 ||∞ ≤ ||un ||∞
assume λ >
1
2
ok
, we will show that the scheme is unstable
consider u0j = (−1)j
u1j = λu0j+1 + (1 − 2λ)u0j + λu0j−1 = λ(−1)j+1 + (1 − 2λ)(−1)j + λ(−1)j−1
= (−1)j (−λ + (1 − 2λ) − λ) = (1 − 4λ) · u0j
λ>
1
2
⇒ 1 − 4λ < −1 ⇒ ||u1 ||∞ > ||u0 ||∞
ok
43
ex : un+1
= unj + kD+ D− unj , h = 0.05 , λ =
j
1
2
⇒ kc = λh2 = 0.00125
t=0
t=0
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
0.2
0.4
0.6
0.8
1
0
0
h=0.05 , k=0.0013 , 1 timestep
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0.2
0.4
0.6
0.8
1
0
0
h=0.05 , k=0.0013 , 25 timesteps
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0.2
0.4
0.6
0.8
1
0
0
h=0.05 , k=0.0013 , 50 timesteps
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0.2
0.4
0.6
0.8
0.8
1
0.2
0.4
0.6
0.8
1
0.2
0.4
0.6
0.8
1
h=0.05 , k=0.0012 , 50 timesteps
1
0
0.6
h=0.05 , k=0.0012 , 25 timesteps
1
0
0.4
h=0.05 , k=0.0012 , 1 timestep
1
0
0.2
1
0
0
0.2
0.4
0.6
0.8
1
13. Thurs 2/17
44
convergence
v(x, t) : exact solution
vt = vxx , 0 ≤ x ≤ 1 , v(x, 0) = f (x) , v(0, t) = v(1, t) = 0
unj : numerical solution
un+1
− unj
unj+1 − 2unj + unj−1
j
=
k
h2
h=
1
N
j = 1, . . . , N − 1
,
, xj = jh , j = 0, . . . , N
tn = nk , k = 0, 1, 2, . . .
u0j = f (xj ) ,
un0 = unN = 0
claim
Assume that x = xj = jh, t = tn = nk and λ = k/h2 are fixed. If λ ≤ 1/2, then
lim unj = v(x, t).
h→0
pf
1. local truncation error
n
n
vjn+1 − vjn
vj+1
− 2vjn + vj−1
=
+ τjn
2
k
h
(contrast to ODEs)
let v = vjn , vt = (vt )nj , . . .
+ O(k 3 )
vjn+1 = v + kvt +
k2
2 vtt
n
vj+1
= v + hvx +
h2
2 vxx
+
h3
6 vxxx
+
h4
24 vxxxx
+ O(h5 )
n
vj−1
= v − hvx +
h2
2 vxx
−
h3
6 vxxx
*
+
h4
24 vxxxx
+ O(h5 )
τjn = vt + k2 vtt + O(k 2 ) − vxx +
h2
12 vxxxx
+
+ O(h4 )
vt = vxx , vtt = vxxt = vtxx = vxxxx
τjn
=
*
k
2
−
h2
12
+
vxxxx + O(k 2 ) + O(h4 )
τjn = O(k) + O(h2 ) :
1st order in time , 2nd order in space
45
2. error bound
un+1
= λunj+1 + (1 − 2λ)unj + λunj−1
j
n
n
vjn+1 = λvj+1
+ (1 − 2λ)vjn + λvj−1
+ kτjn
define enj = vjn − unj , en+1
= λenj+1 + (1 − 2λ)enj + λenj−1 + kτjn
j
λ ≤ 1/2 ⇒ ||en+1 ||∞ ≤ ||en ||∞ + k||τ ||∞ ≤ ||en−1 ||∞ + 2k||τ ||∞
· · · ||en ||∞ ≤ ||e0 ||∞ + nk||τ ||∞ = t · O(k)
ok
note
1. We see (again) that for a consistent scheme, stability ⇒ convergence.
2. unj = v(x, t) + O(k) = v(x, t) + kEjn + O(k 2 ) : hw
alternative : method of lines
vt = vxx , 0 ≤ x ≤ 1 , v(x, 0) = f (x) , v(0, t) = v(1, t) = 0
define uj (t) ≈ v(xj , t) , xj = jh , h = 1/N , j = 1, . . . , N − 1
t
0
u!j =
xj−1
xj
xj+1
1
x
uj+1 − 2uj + uj−1
: system of ODEs , u0 = uN = 0
h2

u! = Au ,
u =









u1
..
.
..
.
..
.
uN −1











,
−2
1

 1 −2
1 

...
A = 2

h 


A : tridiagonal , symmetric ⇒ real e-values
1
...
...

...
...
1







1

−2
Euler’s method : un+1 = un + kAun : finite-difference scheme
46
recall : absolute stability ⇔ − 2 ≤ kµ ≤ 0 for all e-values µ of A
thm (Gershgorin)
If µ is an e-value of A = (aij ), then there exists i st |µ − aii | ≤
pf : Math 571 (or exercise)
%
%
%
%µ
%
!
"%
%
%
%
%
−4
≤ µ ≤ 0
h2
!
"
−4
hence, absolute stability holds if −2 ≤ k
≤ 0 ⇔
h2
G ⇒
−2
−
h2
≤
2
h2
$
j+=i
|aij |.
⇒
k
1
≤
h2
2
ex : compute e-values and e-vectors of A
Au = µu ,
u = (u1 , . . . , uN −1 )T
uj+1 − 2uj + uj−1
= µuj , u0 = uN = 0
h2
uj+1 − (2 + µh2 )uj + uj−1 = 0
:
difference equation
ρ(ζ) = ζ 2 − (2 + µh2 )ζ + 1 = (ζ − ζ1 )(ζ − ζ2 ) = 0
uj = a1 ζ1j + a2 ζ2j
u0 = 0 ⇒ a1 + a2 = 0 ⇒ a2 = −a1
uN = 0 ⇒
a1 (ζ1N
−
ζ1
= e2πim/N = e2πimh
ζ2
ζ2N )
=0 ⇒
ζ1N
=
ζ2N
, m = 1, . . . , N − 1
⇒
!
ζ1
ζ2
"N
=1
(m = 0 gives a contradiction)
ζ1 ζ2 = 1 ⇒ ζ12 = e2πimh ⇒ ζ1 = eπimh , ζ2 = e−πimh since ζ1 , ζ2 → 1 as h → 0
9
:
uj = a1 ζ1j + a2 ζ2j = a1 eπijmh − e−πijmh = a1 · 2i sin πjmh
ζ1 + ζ2 − 2
eπimh + e−πimh − 2
2(cos πmh − 1)
ζ1 +ζ2 = 2+µh ⇒ µ =
=
=
2
2
h
h
h2
−4
πmh
e-values of A : µm = 2 sin2
, m = 1, . . . , N − 1
h
2
e-vectors of A : um,j = sin πmxj , j = 1, . . . , N −1 : discrete Fourier modes
2π
2
πm = ξ : wavenumber , wavelength =
=
ξ
m
2
ex : N = 16 , e-values and e-vectors of A = D+ D− + Dirichlet BC
!
0
N
!4 / h2
m=1
m=2
m=4
m=15
note : m → 0 : long wavelength modes ⇒ µ → 0
m → N : short · · · · · · · ” · · · · · · · · ⇒ µ → −4/h2
m
47
14. Tues 2/22
48
matrix analysis of convergence
vt = vxx
un+1 = un + kAun : method of lines + Euler’s method
un+1 = (I + kA)un
v n+1 = (I + kA)v n + kτ n
en = v n − un ⇒ en+1 = (I + kA)en + kτ n
||en+1 || ≤ ||I + kA|| · ||en || + k||τ n ||

1 − 2λ
λ


λ
1 − 2λ


...

I + kA = 



||I + kA||∞ = ||I + kA||1 =
λ
...
...
...
...
λ
7

λ
1 − 2λ









,
λ=
k
h2
1 if λ ≤ 1/2
4λ − 1 if λ > 1/2
||I + kA||2 = max of absolute values of e-values of I + kA
Gershgorin ⇒ the e-values of A lie in the interval [−4/h2 , 0]
1
2
hence λ ≤
⇒ · · · · · · · · · · · I + kA · · · · · · · · · · · · · [1 − 4λ , 1]
⇒ ||I + kA||p ≤ 1 for p = 1, 2, ∞
||en+1 || ≤ ||en || + k||τ n ||
||en || ≤ ||e0 || + t||τ || , ||τ || = max
||τ n ||
n
ok
Fourier analysis of PDE
vt = vxx
look for solutions of the form v(x, t) = eωt+iξx : Fourier mode
ξ : wavenumber , ω : growth rate , ω = −ξ 2 : dispersion relation
−ξ 2 t+iξx
v(x, t) = e
7
ξ = 0 : constant mode
ξ += 0 : oscillatory in space, decaying in time
49
IVP with free-space BC : PDE
−∞<x<∞
vt = vxx , v(x, 0) = f (x) ,
tools
Fourier transform : f?(ξ) =
inversion formula : f (x) =
8 ∞
Parseval’s relation :
claim
1. v(x, t) =
8 ∞
−∞
−∞
f?(ξ)e−ξ
2
8 ∞
−iξx
1
dx
2π −∞ f (x)e
8 ∞
−∞
f?(ξ)eiξx dξ
|f (x)|2 dx = 2π
t+iξx
8 ∞
−∞
|f?(ξ)|2 dξ
,
||f ||2 = 2π||f?||2
dξ : solution formula
2. ||v(·, t)||2 ≤ ||f ||2 for all t ≥ 0 : L2 -stability
pf
1.
?
v(ξ,
t)
v?t (ξ, t)
=
8 ∞
−iξx
1
dx
2π −∞ v(x, t)e
8 ∞
−iξx
1
dx
2π −∞ vt (x, t)e
=
%∞
−iξx %%
vx (x, t)e
%
−∞
=
1
2π
=
1
− 2π
8 ∞
−iξx
1
dx
2π −∞ vxx (x, t)e
=
−
8 ∞
−iξx
1
dx
2π −∞ vx (x, t)(−iξ)e
%∞
−iξx %%
v(x, t)(−iξ)e
%
−∞
8 ∞
2 −iξx
1
dx
2π −∞ v(x, t)(−iξ) e
+
?
?
v?t (ξ, t) = −ξ 2 v(ξ,
t) , v(ξ,
0) = f?(ξ)
?
⇒ v(ξ,
t) = f?(ξ)e−ξ
v(x, t) =
8 ∞
−∞
2
t
?
v(ξ,
t)eiξx dξ =
2. ||v(·, t)||22 =
8 ∞
−∞
= 2π
≤ 2π
8 ∞
−∞
2
f?(ξ)e−ξ t eiξx dξ
|v(x, t)|2 dx = 2π
8 ∞
−∞
|f?(ξ)|2 e−2ξ t dξ
−∞
|f?(ξ)|2 dξ =
8 ∞
8 ∞
−∞
2
8 ∞
−∞
ok
?
|v(ξ,
t)|2 dξ
|f (x)|2 dx = ||f ||22
ok
50
Fourier analysis of difference scheme
9
un+1
= unj + λ unj+1 − 2unj + unj−1
j
:
look for solutions of the form unj = ζ n eiξjh
9
ζ n+1 eiξjh = ζ n eiξjh + λ ζ n eiξ(j+1)h − 2ζ n eiξjh + ζ n eiξ(j−1)h
:
ζ = 1 + λ(eiξh − 2 + e−iξh ) = 1 + 2λ(cos ξh − 1) = 1 − 4λ sin2 ξh
2 = ρ(ξh)
un+1
= ρ(ξh)unj , ρ(ξh) : amplification factor
j
ρ(ξh)
1
λ=0
0
λ = 1/4
−1
λ = 1/2
λ > 1/2
π/2
π
note
1. |ρ(ξh)| ≤ 1 for all ξh ⇔ λ ≤
1
2
2. unj = ρ(ξh)n u0j
0≤λ≤
1
4
1
4
1
2
≤λ≤
λ>
1
2
: all modes decay monotonically in time
:
:
7
long waves decay monotonically in time
short waves oscillate in sign, amplitude decays




long waves decay monotonically in time
intermediate waves oscillate in sign, amplitude decays
short waves oscillate in sign, amplitude grows



3. amplification factor of PDE = e−ξ k = e−λ(ξh) = ρ(ξh) + O((ξh)4 )
For the PDE, all modes decay monotonically in time, in contrast with the difference scheme.
2
2
15. Tues 3/8
51
IVP with free-space BC : difference scheme
un+1
= unj + kD+ D− unj , u0j : given , j = 0, ±1, . . .
j
goal : derive results for unj analogous to results for v(x, t)
tools
Fourier coefficients :
f?m
=
8 π
1
2π −π
f (x)e−imx dx
m=−∞
f?m eimx
inversion formula : f (x) =
Parseval’s relation :
8 π
−π
∞
$
|f (x)|2 dx = 2π
∞
$
m=−∞
|f?m |2
change notation : m → j , f?m → unj , x → ξh , f (x) → u? n (ξh)
unj =
8 π
−π
8 π
−ijξh
1
?n
d(ξh)
2π −π u (ξh)e
|u? n (ξh)|2 d(ξh)
= 2π
∞
$
j=−∞
, u? n (ξh) =
∞
$
j=−∞
unj eijξh
|unj |2
claim
1.
unj
=
2. λ ≤
1
2
8 π
n −ijξh
1
?0
d(ξh)
2π −π u (ξh)ρ(ξh) e
: solution formula
⇒ ||un ||2 ≤ ||u0 ||2 for all n ≥ 0 : stability
pf
1. u? n+1 (ξh) =
∞
$
j=−∞
un+1
eijξh =
j
∞
$
(unj + λ(unj+1 − 2unj + unj−1 ))eijξh
j=−∞
= (1 + λ(e−iξh − 2 + eiξh ))u? n (ξh) = ρ(ξh)u? n (ξh)
⇒ u? n (ξh) = ρ(ξh)n u? 0 (ξh)
2. ||un ||22 =
≤
∞
$
j=−∞
|unj |2 =
ok
8 π
2
1
?n
2π −π |u (ξh)| d(ξh)
8 π
2
1
?0
2π −π |u (ξh)| d(ξh)
=
∞
$
j=−∞
=
8 π
1
2π −π
|ρ(ξh)n u? 0 (ξh)|2 d(ξh)
|u0j |2 = ||u0 ||22
ok
52
implicit schemes
PDE : vt = vxx , 0 ≤ x ≤ 1 , v(0, t) = v(1, t) = 0
method of lines : u! = Au , A =
1
h2
trid(1, −2, 1)
the e-values of A lie in the interval (− h42 , 0) ⇒ stiff system
forward Euler : conditionally stable , i.e. stable ⇔ λ ≤
1
2
backward Euler : unconditionally stable , i.e. stable for all λ > 0 (hw)
Crank-Nicolson : centered differencing in space + trapezoid in time
un+1 = un + k2 (Aun + Aun+1 ) : error is O(h2 ) + O(k 2 ) , 2nd order accurate
(I − k2 A)un+1 = (I + k2 A)un : linear system (more later)
claim : CN is unconditionally stable and convergent
pf
1. un+1 = (I − k2 A)−1 (I + k2 A)un ⇒ ||un+1 ||2 ≤ ||(I − k2 A)−1 ||2 · ||I + k2 A||2 · ||un ||2
the e-values of I − k2 A lie in the interval (1, 1 + 2λ)
· · · · · · ” · · · · · I + k2 A · · · · · · · ” · · · · · · (1 − 2λ, 1)
||(I − k2 A)−1 ||2 ≤ 1 , ||I + k2 A||2 ≤ max{1, |1 − 2λ|} ≤ 1 ⇔ λ ≤ 1
then λ ≤ 1 ⇒ ||un+1 ||2 ≤ ||un ||2 : not sharp
1 + k2 µ
if µ is an e-value of A , then
is an e-value of (I − k2 A)−1 (I + k2 A)
k
1 − 2µ
µ < 0 ⇒ ||(I − k2 A)−1 (I + k2 A)||2 ≤ 1 ⇒ ||un+1 ||2 ≤ ||un ||2 for all λ > 0
2. un+1 = un + k2 (Aun + Aun+1 )
v n+1 = v n + k2 (Av n + Av n+1 ) + kτ n
, τ n = O(k 2 )
en = v n − un ⇒ en+1 = en + k2 (Aen + Aen+1 ) + kτ n
en+1 = (I − k2 A)−1 (I + k2 A)en + (I − k2 A)−1 kτ n
||en+1 ||2 ≤ ||en ||2 + k||τ n ||2 ⇒ ||en ||2 ≤ ||e0 ||2 + t · O(k 2 )
ok
ok
16. Thurs 3/10
53
Fourier analysis (Crank-Nicolson)
un+1
= unj + k2 (D+ D− unj + D+ D− un+1
)
j
j
n+1
λ n
n
n
n
un+1
− λ2 (un+1
+ un+1
j
j+1 − 2uj
j−1 ) = uj + 2 (uj+1 − 2uj + uj−1 )
look for solutions of the form unj = ζ n eiξjh
1 + λ2 (eiξh − 2 + e−iξh )
1 − 2λ sin2 ξh
2 = ρ(ξh) : amplification factor
ζ =
=
λ iξh
2 ξh
−iξh
1 − 2 (e − 2 + e
)
1 + 2λ sin 2
unj = ρ(ξh)n u0j
ρ(ξh)
1
λ=0
λ = 1/4
0
λ = 1/2
λ=1
λ>1
−1
π/2
note
1. 0 ≤ λ ≤
1
2
λ>
1
2
π
: all modes decay monotonically in time
:
7
long waves decay monotonically in time
short waves oscillate in sign, amplitude decays
2. |ρ(ξh)| ≤ 1 for all ξh and all λ > 0
⇒ CN is unconditionally stable (for the IVP with free-space BC)
un+1
= unj + k2 (D+ D− unj + D+ D− un+1
) , j = 0, ±1, . . .
j
j
unj
n
=
8 π
1
2π −π
ρ(ξh)n u? 0 (ξh)e−ijξh d(ξh)
0
||u ||2 ≤ ||u ||2 for all n ≥ 0 , all λ > 0







: pf as before
54
recall : (I − k2 A)un+1 = (I + k2 A)un
write this in the form Ax = f
tridiagonal Gaussian elimination : special case

a

b


A =




b
a
...
b
...
...
...
...
b

b
a










1

β
 2

=




1
β3

1
...
...
βn
1









α1
b
α2
b
...
...
...

b
αn









A = LU , L : unit lower triangular , U : upper triangular
steps
1. find L, U
2. solve Ly = f
3. solve U x = y
check Ax = LU x = Ly = f
ok
find L, U
a = α1 , b = βk αk−1 ⇒ βk =
b
αk−1
, a = βk b + αk ⇒ αk = a − βk b , k = 2 : n
solve Ly = f
y1 = f1 , fk = βk yk−1 + yk ⇒ yk = fk − βk yk−1 , k = 2 : n
solve U x = y
yn = αn xn ⇒ xn =
yn
yk − bxk+1
, yk = αk xk + bxk+1 ⇒ xk =
, k = n−1 : 1
αn
αk
note
the operation count is O(n)
17. Tues 3/15
55
summary on Fourier analysis : we are using 4 types of transforms
1 continuous /continuous f (x)/f?(ξ)
−∞ < x < ∞ / − ∞ < ξ < ∞
3 discrete/continuous
j = 0, ±1, . . . / − π ≤ ξh ≤ π
f (x)/{f?m }
2 continuous /discrete
?
{uj }/u(ξh)
4 discrete/discrete
{uj }/{u? m }
0 ≤ x ≤ 1 / m = 0, ±1, . . .
j = 1 : N − 1/m = 1 : N − 1
In each case there is an inversion formula and a Parseval’s relation. The specific
form of the transform depends on the boundary conditions. Fourier analysis is
used to prove stability in the 2-norm for PDEs and difference schemes.
ex : stability analysis of a difference scheme using a discrete/discrete transform
vt = vxx , 0 ≤ x ≤ 1 , v(0, t) = v(1, t) = 0
un+1 = (I+kA)un , un = {unj } , un0 = unN = 0 , A =
e-values of A : µm =
−4
h2
sin2 πmh
, m=1:N −1
2
1
h2 trid(1, −2, 1)
, Nh = 1
e-vectors of A : qm = {sin πmjh} , j = 1 : N − 1 , orthonormal basis
any vector u = {uj } can be expanded in this basis
u =
N$
−1
m=1
u? m qm ⇒ uj =
u? m = u · qm =
N$
−1
N$
−1
m=1
u? m sin πmjh : inversion formula
uj sin πmjh : discrete Sine transform
j=1
u? = {u? m } : the components of u? are the cofficients of u wrt the basis {qm }
||u||22 =
N$
−1
j=1
u2j = u · u =
N$
−1
m=1
? 2 : Parseval’s relation
u? 2m = ||u||
2
now consider the difference scheme
0
u =
N$
−1
m=1
u? 0m qm
,
n
n 0
u = (I + kA) u =
N$
−1
m=1
|1 + kµm | = |1 − 4λ sin2 πmh
2 | ≤ 1 if λ ≤ 1/2
u? 0m (1 + kµm )n qm
then ||un ||2 = ||u? n ||2 ≤ ||u? 0 ||2 = ||u0 ||2 : stability
hw : Neumann BC , vx (0, t) = vx (1, t) = 0
56
energy method : alternative technique for proving stability in the 2-norm
vt = vxx , − ∞ < x < ∞
*8 ∞
define ||v(·, t)||2 =
−∞
( ok too for 0 ≤ x ≤ 1 + Dirichlet BC )
2
v(x, t) dx
+1/2
: energy
claim : ||v(·, t)||2 ≤ ||v(·, 0)||2 for all t ≥ 0 : stability
pf
d
d 8∞ 2
2
||v(·, t)||2 =
v dx =
dt
dt −∞
=
%∞
%
2vvx %% −
−∞
2
8 ∞
8 ∞
−∞
−∞
8 ∞
2vvt dx = 2
(vx )2 dx ≤ 0
−∞
vvxx dx
ok
integration by parts
define (f, g) =
8 ∞
−∞
f (x)g(x) dx : inner product , (f, f ) = ||f ||22
claim : (f, g ! ) = −(f ! , g)
pf
(f g)! = f ! g + f g !
8 ∞
!
(f g) dx =
−∞
8 ∞
−∞
!
!
!
!
(f g + f g ) dx = (f , g) + (f, g ) =
summation by parts
define (f, g) =
∞
$
fj gj
j=−∞
%∞
%
f g %%
−∞
= 0
ok
, (f, f ) = ||f ||22,h
claim : (f, D− g) = −(D+ f, g)
pf
D+ (f g)j =
fj+1 gj+1 − fj gj
fj+1 gj+1 − fj+1 gj + fj+1 gj − fj gj
=
h
h
= fj+1 D+ gj + (D+ fj )gj
∞
$
D+ (f g)j =
j=−∞
∞
$
(fj+1 D+ gj + (D+ fj )gj ) =
j=−∞
= (f, D− g) + (D+ f, g) =
∞
$
fj D− gj +
j=−∞
∞
$
fj+1 gj+1 − fj gj
= 0
h
j=−∞
ok
∞
$
(D+ fj )gj
j=−∞
57
application to difference schemes
n
||u ||2 =
!
∞
$
(unj )2
j=−∞
"1/2
: discrete energy
1. forward Euler/centered difference
un+1 = un + kD+ D− un
claim : λ ≤ 1/2 ⇒ ||un ||2 ≤ ||u0 ||2 : conditional stability
pf
||un+1 ||2 − ||un ||2 = (un+1 , un+1 ) − (un , un ) = (un+1 + un , un+1 − un )
= (2un + kD+ D− un , kD+ D− un ) = 2k(un , D+ D− un ) + k 2 ||D+ D− un ||2
(1) = −2k(D− un , D− un ) = −2k||D− un ||2
(2) =
%%
n
%%
2 %% S+ D− u −
k %%%%
h
%%

D− un %%%%2
%%
%%
≤
k2
k2
n
n 2
(||S
D
u
||
+
||D
u
||)
=
· 4||D− un ||2
+ −
−
2
2
h
h

k2
||un+1 ||2 − ||un ||2 ≤ −2k + 4 2 ||D− un ||2 = 2k(2λ − 1)||D− un ||2 ≤ 0
h
ok
2. backward Euler/centered difference
un+1 = un + kD+ D− un+1 : unconditionally stable (hw)
3. Crank-Nicolson
9
un+1 = un + k2 D+ D− un + un+1
pf
:
: unconditionally stable
||un+1 ||2 − ||un ||2 = (un+1 + un , un+1 − un ) = (un+1 + un , k2 D+ D− (un + un+1 ))
2D heat equation
(chap. 3)
vt = vxx + vyy , v(x, y, 0) = f (x, y) ,
unj,l ≈ v(jh, lh, nk)
x n
D+
uj,l
y n
D+
uj,l
= − k2 ||D− (un +un+1 )||2 ≤ 0
− ∞ < x, y < ∞
unj+1,l − unj,l
unj,l − unj−1,l
x n
=
, D− uj,l =
h
h
unj,l+1 − unj,l
unj,l − unj,l−1
y n
=
, D− uj,l =
h
h
ok
18. Thurs 3/17
58
forward Euler/centered difference
y
y
x x
un+1
= unj,l + k (D+
D− + D+
D−
) unj,l
j,l
9
un+1
= unj,l + λ unj+1,l − 2unj,l + unj−1,l + unj,l+1 − unj,l + unj,l−1
j,l
note
9
1. un+1
= (1 − 4λ)unj,l + λ unj+1,l + unj−1,l + unj,l+1 + unj,l−1
j,l
:
:
claim : ||un ||∞ ≤ ||u0 ||∞ ⇔ λ ≤ 1/4 : maximum principle
pf as before
2. unj,l = ρ(ξ, η)n ei(ξj+ηl)h : special solutions
!
ξh
ηh
ρ(ξ, η) = 1 + λ (2 cos ξh − 2 + 2 cos ηh − 2) = 1 − 4λ sin
+ sin2
2
2
|ρ(ξ, η)| ≤ 1 ⇔ λ ≤ 1/4
unj,l =
1 8π 8π n
u? (ξh, ηh)ei(ξh+ηl)h d(ξh)d(ηh)
2
−π
−π
(2π)
u? n (ξh, ηh) =
∞
$
j,l=−∞
|unj,l |2
∞
$
j,l=−∞
unj,l e−i(ξj+ηl)h
1 8π 8π n
=
|u? (ξh, ηh)|2 d(ξh)d(ηh)
2
(2π) −π −π
claim : λ ≤ 1/4 ⇒ ||un ||2 ≤ ||u0 ||2
pf : u? n+1 (ξh, ηh) = ρ(ξh, ηh)u? n (ξh, ηh) . . . as before
backward Euler/centered difference
y
y
x x
un+1 = un + k (D+
D− + D+
D−
) un+1
y
y
x x
(I − k (D+
D− + D+
D−
)) un+1 = un : linear system
9
:
n+1
n+1
n+1
n+1
n
(1 + 4λ)un+1
j,l − λ uj+1,l + uj−1,l + uj,l+1 + uj,l−1 = uj,l
consider the problem on D = [0, 1]2
v(x, y, 0) = f (x, y) on D
v(x, y, t) = 0 on ∂D
2
"
59
y
4
3
3
6
9
2
2
5
8
1
1
4
7
x
0
0
1
2
3
4
1
2
3
4
5
6
7
8
9
u1,1
u1,2
u1,3
u2,1
u2,2
u2,3
u3,1
u3,2
u3,3
1 + 4λ
−λ
−λ
−λ
1 + 4λ
−λ
−λ
−λ
1 + 4λ
−λ
−λ
1 + 4λ
−λ
−λ
−λ
−λ
1 + 4λ
−λ
−λ
−λ
−λ
1 + 4λ
−λ
−λ
1 + 4λ
−λ
−λ
−λ
1 + 4λ
−λ
−λ
−λ
1 + 4λ
note
1. the matrix is symmetric, block tridiagonal, positive definite : Math 571
Cholesky, conjugate gradient, FFT, SOR, multigrid, block LU, . . .
2. ρ(ξh, ηh) = ? : 1D case is on hw
|ρ(ξh, ηh)| ≤ 1 for all λ : unconditional stability in 2-norm
(can also be shown by energy method)
3. global error is O(k) + O(h2 )
19. Tues 3/22
60
operator splitting
y ! = Ay ⇒ y(t) = eAt y0
eAt = I + At + 12 A2 t2 + · · ·
define SA (t) = eAt : solution operator
y(t) = SA (t)y0
9
SA (t) = eAt = eAnk = eAk
consider y ! = (A + B)y
:n
= (SA (k))n
claim
SA+B (k) = e(A+B)k += eAk eBk = SA (k)SB (k)
pf
SA+B (k) = e(A+B)k = I + (A + B)k + 12 (A + B)2 k 2 + · · ·
(A + B)2 = (A + B)(A + B) = A2 + AB + BA + B 2
SA+B (k) = I + (A + B)k + 12 (A2 + AB + BA + B 2 )k 2 + · · ·
SA (k)SB (k) = eAk eBk = (I + Ak + 12 (Ak)2 + · · ·)(I + Bk + 12 (Bk)2 + · · ·)
= I + (A + B)k + ( 12 A2 + AB + 12 B 2 )k 2 + · · ·
SA+B (k) − SA (k)SB (k) = 12 (BA − AB)k 2 + · · ·
ok
SA+B (k) = SA (k)SB (k) + O(k 2 )
SA+B (nk) = (SA+B (k))n = (SA (k)SB (k) + O(k 2 ))n = (SA (k)SB (k))n + O(k)
note : this holds more generally , 1st order accurate in time
application : backward Euler/centered difference scheme for 2D heat equation
y
y
x x
un+1 = un + k(D+
D− + D+
D−
)un+1
y
y
x x
(I − k(D+
D− + D+
D−
))un+1 = un
y
y −1
x x
un+1 = SA+B (k)un , SA+B (k) = (I − k(D+
D− + D+
D−
))
y
y −1
x x −1
set SA (k) = (I − kD+
D− ) , SB (k) = (I − kD+
D−
)
61
claim : SA+B (k) = SA (k)SB (k) + O(k 2 )
pf
y
y
x x
un = (I − k(D+
D− + D+
D−
))un+1
y
y
y
y
x x
x x n+1
= (I − kD+
D−
)(I − kD+
D− )un+1 − k 2 D+
D−
D+
D− u
y −1 n
y
x x −1
un+1 = (I − kD+
D− ) (I − kD+
D−
) u + O(k 2 )
ok
The backward Euler/centered difference scheme can be replaced by the following.
1
y
y
(I − kD+
D−
)un+ 2 = un
x x
(I − kD+
D− )un+1 = u
note
n+ 12







: fractional-step method
Only 1D tridiagonal systems need to be solved (if the vector components are
ordered correctly).
1st half-step : u1 , u2 , u3 , u4 , u5 , u6 , u7 , u8 , u9
2nd half-step : u1 , u4 , u7 , u2 , u5 , u8 , u3 , u6 , u9
unconditionally stable in 2-norm , global error is O(k) + O(h2 )
pf . . .
Crank-Nicolson
9
y
y
x x
un+1 = un + k2 (D+
D− + D+
D−
) un + un+1
:
y
y
y
y
x x
x x
(I − k2 (D+
D− + D+
D−
))un+1 = (I + k2 (D+
D− + D+
D−
))un
block-tridiagonal, symmetric, positive definite
9
:
1 − 2λ sin2 ξh
+ sin2 ηh
2
2 :
9
ρ(ξh, ηh) =
2 ξh
2 ηh
1 + 2λ sin 2 + sin 2
⇒ |ρ(ξh, ηh)| ≤ 1 for all ξ , η , λ
unconditionally stable in 2-norm , global error is O(k 2 ) + O(h2 )
fractional-step/Crank-Nicolson
y
y
y
y
x x
(I − k2 D+
D−
)un+1/2 = (I + k2 (D+
D− + D+
D−
))un
x x
(I − k2 D+
D− )un+1
= un+1/2
tridiagonal systems, unconditionally stable, O(k) + O(h2 )
pf . . .
62
alternating direction implicit
y
y
x x
(I − k2 D+
D−
)un+1/2 = (I + k2 D+
D− )un
y
y
= (I + k2 D+
D−
)un+1/2
x x
(I − k2 D+
D− )un+1
tridiagonal systems
stability of ADI
1 − 2λ sin2 ξh
2
ρ1 (ξh, ηh) =
2 ηh
1 + 2λ sin 2
,
1 − 2λ sin2 ηh
2
ρ2 (ξh, ηh) =
2 ξh
1 + 2λ sin 2
ρ1 (π, 0) = ρ2 (0, π) = 1 − 2λ
⇒ each fractional step is conditionally stable : λ ≤ 1
1 − 2λ sin2 ξh
1 − 2λ sin2 ηh
2
2
ρ1 (ξh, ηh) = ρ1 (ξh, ηh) · ρ2 (ξh, ηh) =
·
2 ηh
2 ξh
1 + 2λ sin 2 1 + 2λ sin 2
⇒ |ρ(ξh, ηh)| ≤ 1 for all ξ , η , λ : each full step is unconditionally stable
accuracy of ADI
k
2
9
un+1 − un+1/2 =
k
2
un+1/2 − un =
y
y n+1/2
x x n
D+
D−
u
+ D+
D− u
9
:
y
y n+1/2
x x n+1
D+
D− u
+ D+
D−
u
9
:
:
y
y n+1/2
x x
un+1 − un = k2 D+
D− un+1 + un + kD+
D−
u
x x
2un+1/2 − (un+1 + un ) = k2 D+
D− (un − un+1 )
x x
un+1/2 = 12 (un+1 + un ) − k4 D+
D− (un+1 − un )
un+1 − un =
&
k
2
9
:
y
y
x x
(D+
D− + D+
D−
) un+1 + un −
'(
Crank-Nicolson
)
⇒ the global error in ADI is O(h2 ) + O(k 2 )
y
k2 y
x x
4 D+ D− D+ D−
&
'(
3
k
4
9
un+1 − un
vxxyyt
:
)
20. Thurs 3/24
63
hyperbolic equations
wave propagation : sound , light , water , gravity
scalar wave equation
vt + cvx = 0 ,
− ∞ < x < ∞ , v(x, 0) = f (x)
(assume c > 0 wlog)
def
The line x − ct = α in the xt-plane is called a characteristic.
t
x − ct = α , slope =
α
1
c
x
claim
1. The solution v(x, t) is constant on characteristics.
2. The solution is v(x, t) = f (x − ct).
pf
1. v(x, t) = v(α + ct, t)
d
v(α + ct, t) = vx (α + ct, t) · c + vt (α + ct, t) = 0
dt
2. v(x, t) = v(α, 0) = f (α) = f (x − ct)
ok
ok
check : if v(x, t) = f (x − ct), then vt + cvx = f ! · −c + c · f ! = 0
note
1. The solution is a traveling wave with wave speed c.
2. The domain of dependence of the differential equation is {α}.
ok
64
linear system of 1st order PDEs


v1
 . 
. 
vt + Avx = 0 , v(x, 0) = f (x) , v = 
 .  , f =
vp


f1
 . 
 .. 


fp
The system is called hyperbolic if A is diagonalizable and has real e-values,
A = T DT −1 , D = diag(λ1 , . . . , λp ) , λi ∈ IR. The e-values of A are the wave
speeds of the system.
vt + Avx = vt + T DT −1 vx = 0
w = T −1 v ⇒ wt + Dwx ⇒ (wi )t + λi (wi )x , i = 1, . . . , p
ex
!
v1
v2
"
A =
+
t
!
!
0 1
1 0
0 1
1 0
"
"!
v1
v2
"
= T DT
wt + Dwx = 0 ⇒
v1 (x, 0) = f1 (x)
v2 (x, 0) = f2 (x)
= 0 ,
x
!
1
0
, D=
0 −1
−1
"
(w1 )t + (w1 )x = 0
(w2 )t − (w2 )x = 0
!
1 1
1
, T =√
2 1 −1
⇒
w1 (x, 0)
w2 (x, 0)
"
= T
−1
!
f1 (x)
f2 (x)
"
!
1 f1 (x) + f2 (x)
= √
2 f1 (x) − f2 (x)
"
v = Tw
!
v1 (x, t)
v2 (x, t)
"
= T
!
w1 (x, t)
w2 (x, t)
!
"
!
1 w1 (x, t) + w2 (x, t)
= √
2 w1 (x, t) − w2 (x, t)
1 w1 (x − t, 0) + w2 (x + t, 0)
= √
2 w1 (x − t, 0) − w2 (x + t, 0)
"
"
!
1 f1 (x − t) + f2 (x − t) + f1 (x + t) − f2 (x + t)
=
2 f1 (x − t) + f2 (x − t) − f1 (x + t) + f2 (x + t)
The solution is a superposition of traveling waves.
, T = T −1
w1 (x, t) = w1 (x − t, 0)
w2 (x, t) = w2 (x + t, 0)
w = T −1 v ⇒ w(x, 0) = T −1 v(x, 0) = T −1 f (x)
!
"
"
65
ex : 2nd order scalar wave equation
vtt = c2 vxx , v(x, 0) = f (x) , vt (x, 0) = g(x)
v1 = vx ⇒ (v1 )t = vxt = vtx = (v2 )x
(v2 )t = vtt = c2 vxx = c2 (v1 )x
v2 = vt
!
v1
v2
"
+
t
A =
!
D =
!
!
0
−c2
0
−c2
−1
0
c
0
0 −c
"
−1
0
"
"!
v1
v2
"
= 0
x
= T DT −1
, T =
!
1 1
−c c
"
, T
−1
!
1 c −1
=
1
2c c
"
v1 (x, t) =
1
1
(f1 (x + ct) + f1 (x − ct)) +
(f2 (x + ct) − f2 (x − ct))
2
2c
vx (x, t) =
1 !
1
(f (x + ct) + f ! (x − ct)) +
(g(x + ct) − g(x − ct))
2
2c
v(x, t) =
1
1 8 x+ct
(f (x + ct) + f (x − ct)) +
g(s) ds : d’Alembert’s formula
2
2c x−ct
t
(x, t)
x
x − ct
x + ct
The domain of dependence of the differential equation is [x − ct, x + ct].
66
difference schemes
vt + cvx = 0 , v(x, 0) = f (x)
v(x, t) = f (x − ct) , c > 0
upwind
⇒ right-moving wave
downwind
x
downwind scheme
un+1
− unj
j
+ cD+ unj = 0
k
un+1
− unj
un − unj
k
j
+ c j+1
= 0 ⇒ un+1
= (1 + cλ)unj − cλ unj+1 , λ =
j
k
h
h
claim
The downwind scheme is unconditionally unstable in the ∞-norm.
pf
set u0j = (−1)j
then u1j = (1 + cλ)u0j − cλ u0j+1 = (1 + cλ)(−1)j − cλ(−1)j+1 = (−1)j (1 + 2cλ)
||u1 ||∞ = (1 + 2cλ)||u0 ||∞
ok
note
t
x − ct = α
tn
α
xj
x
The domain of dependence of the differential equation is {α}.
The · · · · · · · · · · · · · ” · · · · · · · · · · · · difference scheme is {xj , xj+1 , . . . , xj+n }.
21. Tues 3/29
67
upwind scheme
un+1
− unj
j
+ cD− unj = 0
k
un+1
− unj
un − unj−1
j
+c j
= 0 ⇒ un+1
= (1 − cλ)unj + cλ unj−1
j
k
h
claim
The upwind scheme is stable in the ∞-norm ⇒ cλ ≤ 1.
pf
⇐) un+1
= (1 − cλ)unj + cλ unj−1
j
|un+1
| ≤ (1 − cλ)|unj | + cλ |unj−1 | ≤ (1 − cλ)||un ||∞ + cλ||un ||∞ = ||un ||∞
j
then ||un+1 ||∞ ≤ ||un ||∞
ok
⇒) set u0j = (−1)j
then u1j = (1 − cλ)u0j + cλ u0j−1 = (1 − cλ)(−1)j + cλ(−1)j−1 = (−1)j (1 − 2cλ)
hence if cλ > 1, then ||u1 ||∞ = (2cλ − 1)||u0 ||∞ and the scheme is unstable
cλ < 1 : small k/large h
t
cλ > 1 : large k/small h
t
x − ct = α
tn
α
xj
ok
x − ct = α
tn
x
α
xj
The domain of dependence of the differential equation is {α}.
The · · · · · · · · · · · · ” · · · · · · · · · · · · · difference scheme is {xj−n , . . . , xj }.
x
68
CFL condition : Courant-Friedrichs-Levy
The domain of dependence of the differential equation is contained in the domain
of dependence of the difference scheme.
1. downwind scheme violates CFL for all λ
2. upwind scheme satisfies CFL ⇔ cλ ≤ 1
local truncation error : upwind scheme
vt + cvx = 0
n
vjn − vj−1
vjn+1 − vjn
+c
= τjn
k
h
9
:
v − v − hvx + 12 h2 vxx + O(h3 )
v + kvt + 12 k 2 vtt + O(k 3 ) − v
τ =
+c
k
h
= vt + 12 kvtt + O(k 2 ) + c(vx − 12 hvxx + O(h2 ))
= vt + cvx + 12 h(λvtt − cvxx ) + O(h2 )
vt = −cvx , vtt = −cvxt = −cvtx = −c(−cvx )x = c2 vxx
τ = 12 hc(cλ − 1)vxx + O(h2 ) = O(k)
convergence : upwind scheme , cλ ≤ 1
un+1
= (1 − cλ)unj + cλunj−1
j
n
vjn+1 = (1 − cλ)vjn + cλvj−1
+ kτjn
enj = vjn − unj
en+1
= (1 − cλ)enj + cλenj−1 + kτjn
j
···
||en+1 ||∞ ≤ ||en ||∞ + k||τ n ||∞
···
||en ||∞ ≤ ||e0 ||∞ + nk||τ ||∞ = t · O(k)
⇒ en → 0 if

 u0 = f (xj )
j
 h, k → 0 with
λ=
k
h
fixed st cλ ≤ 1
69
note
The upwind scheme is exact for cλ = 1.
pf
un+1
= (1 − cλ)unj + cλunj−1 = unj−1
j
unj = u0j−n = f (xj−n )
xj−n = xj − nh = xj −
tn
tn
h = xj −
= xj − ctn
k
λ
unj = f (xj−n ) = f (xj − ctn ) = v(xj , tn )
ok
note
Lh unj
un+1
− unj
unj − unj−1
j
=
+c
= 0 , Lh : difference operator
k
h
Lh vjn
n
vjn+1 − vjn
vjn − vj−1
=
+c
= vt + cvx + 12 hc(cλ − 1)vxx + O(h2 )
k
h
vt + cvx = 0 ⇒ Lh vjn = O(h) , unj = vjn + O(h) for cλ ≤ 1
claim
φt + cφx = 12 hc(1 − cλ)φxx ⇒ Lh φnj = O(h2 ) , unj = φnj + O(h2 ) for cλ ≤ 1
&
'(
)
model equation for the upwind scheme ,
1
2 hc(1−cλ)φxx
: artificial viscosity
pf
Lh φnj = O(h2 )
check . . .
enj = φnj − unj , Lh enj = Lh φnj − Lh unj = O(h2 )
···
||en ||∞ ≤ ||e0 ||∞ + t · O(h2 )
ok
note
Stability of the difference scheme is equivalent to requiring ||L−1
h || ≤ 1 for all h
sufficiently small. (more later)
22. Thurs 3/31
70
Fourier analysis : PDE
vt + cvx = 0
look for solutions of the form v(x, t) = eωt+iξx
ω + icξ = 0
⇒
ω = −icξ
v(x, t) = e−icξt+iξx = eiξ(x−ct)
IVP with free-space BC
vt + cvx = 0 , v(x, 0) = f (x) ,
−∞<x<∞
goal : solution formula , stability
?
v(ξ,
t) =
v?t (ξ, t)
1 8∞
v(x, t)e−iξx dx
−∞
2π
1 8∞
=
vt (x, t)e−iξx dx
2π −∞
1 8∞
=
−cvx (x, t)e−iξx dx
2π −∞
%∞
1
1 8∞
−iξx %%
= −c ·
v(x, t)e
+ c·
v(x, t)(−iξ)e−iξx dx
%
−∞
2π
2π −∞
?
?
v?t (ξ, t) = −icξ v(ξ,
t) , v(ξ,
0) = f?(ξ)
?
v(ξ,
t) = f?(ξ)e−icξt
v(x, t) =
8 ∞
−∞
?
v(ξ,
t)eiξx dξ =
=
||v(·, t)||2 = ||f ||2
ok
8 ∞
−∞
f?(ξ)e−icξt eiξx dξ
−∞
f?(ξ)eiξ(x−ct) dξ = f (x − ct)
8 ∞
ok
71
Fourier analysis : difference schemes
look for solutions of the form unj = ρ(ξh)n eijξh
downwind scheme
un+1
= (1 + cλ)unj − cλunj+1
j
⇒
ρ(ξh) = 1 + cλ − cλeiξh
ρ-plane
max |ρ(ξh)| = 1 + 2cλ ⇒
7
unconditionally unstable in 2-norm
short waves are amplified the most
upwind scheme
un+1
= (1 − cλ)unj + cλunj−1
j
0 < cλ <
1
2
⇒
1
2
0 < cλ < 1 ⇒ |ρ(ξh)| ≤ 1 ⇒
cλ > 1 ⇒
ρ(ξh) = 1 − cλ + cλe−iξh
< cλ < 1
7
cλ > 1
scheme is stable in 2-norm
short waves are damped the most
max |ρ(ξh)| = 2cλ − 1 ⇒
7
scheme is unstable in 2-norm
short waves are amplified the most
72
question
In a system with positive and negative wave speeds (e.g. vtt = c2 vxx ), there is
no unique upwind direction. Is there is a scheme which is stable in this case?
centered-difference scheme
vt + cvx = 0
un+1
− unj
unj+1 − unj−1
j
n
n
+ cD0 uj = 0 , D0 uj =
k
2h
un+1
= unj − 12 cλ(unj+1 − unj−1 )
j
claim
1. The CFL condition is satisfied if |c|λ ≤ 1.
2. The scheme is unconditionally unstable in the 2-norm.
pf 1. ok ,
2. ρ(ξh) = 1 − 12 cλ(eiξh − e−iξh ) = 1 − icλ sin ξh
ok
Lax-Friedrichs : add artificial viscosity to stabilize the centered-difference scheme
vt + cvx = 0
un+1
− 12 (unj+1 + unj−1 )
j
+ cD0 unj = 0 ⇒
k
claim
un+1
− unj
k
j
+ cD0 unj =
D+ D− unj
2
k
2λ
1. The CFL condition is satisfied if |c|λ ≤ 1.
2. LF is stable in the 2-norm if |c|λ ≤ 1.
3. LF is 1st order accurate (i.e. error = O(k) ).
4. The model equation for LF is φt + cφx =
h
2λ (1
− c2 λ2 )φxx .
pf : hw
23. Tues 4/5
73
Lax-Wendroff
vt + cvx = 0
⇒
vt = −cvx , vtt = c2 vxx
vjn+1 − vjn
v + kvt + 12 k 2 vtt + O(k 3 ) − v
=
= vt + 12 kvtt + O(k 2 )
k
k
= −cvx + 12 kc2 vxx + O(k 2 )
un+1
− unj
j
+ cD0 unj = 12 kc2 D+ D− unj
k
interpretation : the centered-difference scheme is stabilized by adding artificial
viscosity and the viscosity coefficient is chosen to obtain 2nd order accuracy
claim
1. The CFL condition is satisfied if |c|λ ≤ 1.
2. LW is stable in the 2-norm if |c|λ ≤ 1.
3. LW is 2nd order accurate (i.e. error = O(k 2 ) ).
4. The model equation for LW is (fill in for hw).
pf
un+1
= (1 − c2 λ2 )unj − 12 cλ(1 − cλ)unj+1 + 12 cλ(1 + cλ)unj−1
j
ρ(ξh) = 1 − icλ sin ξh − 2c2 λ2 sin2 ξh
2
= 1 − c2 λ2 + c2 λ2 cos ξh − icλ sin ξh : ellipse
0 < |c|λ <
1
2
1
2
< |c|λ < 1
|c|λ > 1
74
note
If |c|λ ≤ 1, upwind and LF are stable in the ∞-norm, but LW is not.
phase error
compare : v(x, t) = eiξ(x−ct) , unj = ρ(ξh)n eijξh
ρ(ξh) = |ρ(ξh)|eiθ(ξh)
θ(ξh) = arg ρ(ξh) ,
− π < θ(ξh) < π
unj = |ρ(ξh)|n einθ(ξh) eijξh = |ρ(ξh)|n ei(jξh + nθ(ξh))


tn θ(ξh) 
i(jξh + nθ(ξh)) = iξ xj +
= iξ(xj − c# tn )
kξ
#
c# = c(ξh)
=
θ(ξh)
θ(ξh)
=
: numerical wave speed
−kξ
−λ ξh
unj = |ρ(ξh)|n eiξ(xj − #c tn ) : discrete traveling wave (damped or amplified)
def : A system is called dispersive if it has traveling wave solutions in which the
wave speed depends on the wavelength.
1. vt + cvx = 0 is non-dispersive
v(x, t) = eiξ(x − ct) : all waves travel at the same speed
2. φt + φxxx = 0 is dispersive
φ(x, t) = eωt+iξx ⇒ ω + (iξ)3 = 0 ⇒ φ(x, t) = eiξ(x − ξ
2
t)
⇒ c = c(ξ) = −ξ 2 , short waves travel faster than long waves
#
3. c# = c(ξh)
⇒ numerical scheme is dispersive
ex : Lax-Wendroff
ρ(ξh) = 1 − icλ sin ξh − 2c2 λ2 sin2 ξh
2
1. consider long waves , ξh → 0
9
:
ρ(ξh) = 1 − icλ ξh − 16 (ξh)3 + · · · − 2c2 λ2 · 14 (ξh)2 + · · ·
75
−cλξh + 16 cλ(ξh)3 + · · ·
tan θ =
1 − 12 c2 λ2 (ξh)2 + · · ·
9
:9
:
= −cλξh + 16 cλ(ξh)3 + · · · 1 + 12 c2 λ2 (ξh)2 + · · ·
= −cλξh +
9
1
6 cλ
:
− 12 c3 λ3 (ξh)3 + · · ·
= α1 " + α2 "2 + α3 "3 + · · ·
,
" = ξh
θ = β1 " + β2 "2 + β3 "3 + · · ·
tan θ = θ + 13 θ3 + · · ·
9
:
= β1 " + β2 "2 + β3 "3 + · · · +
9
:
1
3
(β1 " + · · ·)3 + · · ·
= β1 " + β2 "2 + β3 + 13 β13 "3 + · · ·
β1 = α1 = −cλ
β2 = α2 = 0
β3 + 13 β13 = α3 ⇒ β3 =
9
1
6 cλ
θ = −cλξh + 16 cλ(1 − c2 λ2 )(ξh)3 + · · ·
c# =
:
− 12 c3 λ3 − 13 (−cλ)3 = 16 cλ(1 − c2 λ2 )
9
:
θ(ξh)
= c 1 − 16 (1 − c2 λ2 )(ξh)2 + · · ·
−λξh
Thus, discrete traveling waves have the correct wave speed in the long wave limit
(ξh → 0), but under the assumption that |c|λ < 1, their wave speed is less than
the exact wave speed, i.e. they exhibit a phase lag.
2. consider short waves , ξh → π , assume c > 0 wlog
ρ(ξh) = 1 − icλ sin ξh − 2c2 λ2 sin2 ξh
2
0 < cλ <
1
2
⇒ ρ(π) = 1 − 2c2 λ2
⇒ 0 < ρ(π) < 1 ⇒ arg ρ(π) = 0 ⇒ c# = 0
⇒ short waves are trapped
1
2
< cλ < 1 ⇒ − 1 < ρ(π) < 0 ⇒ arg ρ(π) = −π ⇒ c# =
⇒ short waves travel faster than the exact wave
note : this explains why LW is not stable in the ∞-norm
1
>c
λ
24. Thurs 4/7
76
hyperbolic systems


v1
 . 

vt + Avx = 0 , v =  .. 

vp
A = T DT −1 , D = diag(c1 , . . . , cp ) , ci : real
w = T −1 v ⇒ wt = Dwx ⇒ (wi )t = ci (wi )x ⇒ wi (x, t) = wi (x − ci t, 0)
characteristics : x − ci t = αi
ex : c1 < c2 < 0 < c3 < c4
t
(x, t)
α4
α3
α2
α1
x
domain of dependence of v(x, t) : {α1 , α2 , α3 , α4 }
Lax-Wendroff :
CFL condition :
un+1
= unj − kAD0 unj + 12 k 2 A2 D+ D− unj
j
max |ci |λ ≤ 1
i
2-norm stability theory for systems
un+1
= Qunj : Q = Q1 S+ + Q0 I + Q−1 S− : 3-point difference operator
j
ex
9
:
QLW = − 12 λA (I − λA) S+ + I − λ2 A2 I + 12 λA (I + λA) S−
Fourier analysis
unj =
8 π
−ijξh
1
?n
d(ξh)
2π −π u (ξh)e
, u? n (mh) =
∞
$
j=−∞
unj eijmh
u? n+1 = G(ξh)u? n , G(ξh) = Q1 eiξh + Q0 + Q−1 e−iξh : amplification matrix
77
u? n (ξh) = Gn (ξh)u? 0 (ξh)
||un ||22
=
∞
$
||unj ||22
j=−∞
=
8 π
1
2π −π
1
≤ max ||Gn (ξh)||22 · 2π
ξh
?n
||u
8 π
−π
(ξh)||22 d(ξh)
=
8 π
1
2π −π
||u? 0 (ξh)||22 d(ξh)
||Gn (ξh)u? 0 (ξh)||22 d(ξh)
||un ||2 ≤ max ||Gn (ξh)||2 · ||u0 ||2
ξh
def
G(ξh) is uniformly power bounded if max ||Gn (ξh)||2 ≤ K for all n ≥ 0, where
ξh
K is independent of h, k, n (but may depend on λ = k/h and t = nk).
note
The scheme un+1
= Qunj is stable in the 2-norm ⇔ G(ξh) is unif power bdd.
j
ex : Lax-Wendroff for hyperbolic systems
9
:
G(ξh) = − 12 λA (I − λA) eiξh + I − λ2 A2 + 12 λA (I + λA) e−iξh
= I − iλA sin ξh − 2λ2 A2 sin2 ξh
2
9
:
−1
= T I − iλD sin ξh − 2λ2 D2 sin2 ξh
2 T
9
Gn (ξh) = T I − iλD sin ξh − 2λ2 D2 sin2 ξh
2
:n
T −1
||Gn (ξh)||2 ≤ ||T ||2 · ||T −1 ||2 if max |ci |λ ≤ 1
i
⇒ G(ξh) is uniformly power bounded and hence LW for a hyperbolic system is
stable in the 2-norm if the CFL condition is satisfied.
ex : leap-frog/centered-difference scheme for scalar wave equation
vt + cvx = 0
un+1
− un−1
j
j
+ cD0 unj = 0
2k
CFL is satisfied if |c|λ ≤ 1. Does this ensure stability in the 2-norm?
!
un+1
j
unj
"
!
−2ckD0
=
1
1
0
"!
unj
un−1
j
"
un+1
= Qunj , Q = Q1 S+ + Q0 I + Q−1 S−
j
78
? n+1
?n
u
iξh
= G(ξh)u
, G(ξh) = Q1 e
−iξh
+ Q0 + Q−1 e
=
!
−2icλ sin ξh 1
1
0
"
det (G(ξh) − µI) = µ2 + 2icλ sin ξh · µ − 1 = 0
µ = −icλ sin ξh ±
@
1 − c2 λ2 sin2 ξh ⇒ |µ| = 1 if |c|λ ≤ 1
µ-plane
i|c|λ
µ2
µ1
-i|c|λ
case 1 : |c|λ < 1
µ1 += µ2 , |µ1 | = |µ2 | = 1 , µ1 + µ2 = −2icλ sin ξh , µ1 µ2 = −1
G(ξh) =
!
µ 1 + µ2
1
!
1
T =
−µ2
1
−µ1
"
1
0
"
= T DT −1
!
µ1
, D=
0
0
µ2
"
, T
−1
1
=
µ2 − µ1
!
−µ1
µ2
−1
1
"
Gn (ξh) = T Dn T −1 ⇒ ||Gn (ξh)||2 ≤ ||T ||2 · ||T −1 ||2
claim
If A is a p × p matrix, then ||A||2 ≤
√
p ||A||∞ and ||A||∞ ≤
√
p ||A||2 .
pf (claim)
1. ||Ax||22 =
⇒
p
$
i=1
|(Ax)i |2 ≤ p ||Ax||2∞ ≤ p ( ||A||∞ · ||x||∞ )2 ≤ p ||A||2∞ · ||x||22
||Ax||2
√
≤ p ||A||∞
||x||2
2. omit
ok
25. Tues 4/12
||T ||∞ ≤ 2
||T −1 ||∞ ≤
√
||T ||2 ≤
79
2
1
1
√
= √
≤
|µ2 − µ1 |
1 − c 2 λ2
1 − c2 λ2 sin2 mh
2·2
√
||T −1 ||2 ≤ 2 · √
⇒
1
1 − c 2 λ2
max ||Gn (ξh)||2 ≤ √
ξh
case 2 : |c|λ = 1
4
: unif power bdd if |c|λ < 1
1 − c 2 λ2
previous bound fails
assume wlog cλ = −1 and consider ξh =
G( π2 )
!
=
!
2i 1
1 0
"
"
=
!
i 1
1 −i
= i
0 i
0 1
Gn ( π2 )
=
!
i 1
1 0
"!
!
i 1
1 0
"
i 1
0 i
"!
i 1
0 i
"!
!
i 1
0 i
"n
⇒
"n !
0 1
1 −i
= i
n
!
"
1 −i
0 1
"
"n
= i
!
n
!
1 −ni
0
1
0 1
n + 1 −ni
= . . . = in
1 −i
−ni 1 − n
max ||Gn (ξh)||2 ≥ ||Gn ( π2 )||2 ≥
ξh
π
2
√1
2
||Gn ( π2 )||∞ =
2n+1
√
2
"
"
: not unif power bdd
Hence the leap-frog/centered-difference scheme for the scalar wave equation is
stable in the 2-norm ⇔ |c|λ < 1. This can be generalized to hyperbolic systems.
note
1. Fourier method ⇒ ||un+1 ||22 + ||un ||22 ≤
2. energy method
pf
⇒ ||un+1 ||22 + ||un ||22 ≤
:
16 9 1 2
0 2
||u
||
+
||u
||
2
2
1 − c 2 λ2
:
1 + |c|λ 9 1 2
||u ||2 + ||u0 ||22
1 − |c|λ
set Ln = ||un ||22 + ||un−1 ||22 + 2kc(un , D0 un−1 ) , show Ln+1 = Ln for n ≥ 1 . . .
80
lower order terms
scalar equation
vt + cvx = bv , v(x, 0) = f (x)
The characteristics are still the lines x − ct = α in the xt-plane.
d
v(α + ct, t) = vx (α + ct, t) · c + vt (α + ct, t) = bv(α + ct, t)
dt
⇒ v(α + ct, t) = ebt v(α, 0) ⇒ v(x, t) = ebt f (x − ct) check . . .
||v(·, t)||2 = ebt ||f ||2 : energy of exact solution changes in time
un+1
= unj − kcD0 unj + 12 k 2 c2 D+ D− unj + kbunj
j
(for example)
ρ(ξh) = ρ0 (ξh) + kb , ρ0 (ξh) : LW
recall : |c|λ ≤ 1 ⇒ |ρ0 (ξh)| ≤ 1 ⇒ |ρ(ξh)| ≤ 1 + k|b|
9
ρn (ξh) ≤ (1 + k|b|)n ≤ ek|b|
:n
= e|b|t , t = nk
u? n (ξh) = ρ(ξh)n u? 0 (ξh) ⇒ ||un ||2 ≤ e|b|t ||u0 ||2 : stability
system
vt + Avx = Bv , A = T DT −1
un+1
= unj − kAD0 unj + 12 k 2 A2 D+ D− unj + kBunj
j
G(ξh) = G0 (ξh) + kB , G0 (ξh) : LW
goal : show that G(ξh) is uniformly power bounded
recall : max |ci |λ ≤ 1 ⇒ ||Gn0 (ξh)||2 ≤ ||T ||2 · ||T −1 ||2 = g , set ||B||2 = b
i
||Gn ||2 = ||(G0 + kB)n ||2 ≤ ||G0 + kB||n2 ≤ (||G0 ||2 + k||B||2 )n = (g + kb)n
= (g(1 + kb/g))n ≤ g n e(b/g)t : ok if g ≤ 1 , not ok if g > 1
case 1 : G0 B = BG0
Gn = (G0 + kB)n =
n 9 :
$
n
i=0
n
||G ||2 ≤
n 9 :
$
n
i=0
n−i
i ||G0 ||2
i
i
Gn−i
0 (kB)
i
· ||(kB) ||2 ≤ g
n 9 :
$
n
i=0
i
(kb)i ≤ g(1 + kb)n ≤ gebt
ok
26. Thurs 4/14
81
case 2 : G0 B += BG0
(G0 + kB)2 = G20 + k(G0 B + BG0 ) + k 2 B 2
(G0 +kB)3 = G30 + k(G20 B +G0 BG0 +BG20 ) + k 2 (G0 B 2 +BG0 B +B 2 G0 ) + k 3 B 3
9 :
n
1 terms with one B and (n − 1) G0
9 :
+ k 2 · n2 terms with two B and (n − 2) G0
9 :
9 :
||(G0 + kB)n ||2 ≤ g + k n1 g 2 b + k 2 n2 g 3 b2 + · · ·
 
n
$
n
g  (kgb)i = g(1 + kgb)n ≤ gegbt
ok
(G0 + kB)n = Gn0 + k ·
||Gn ||2 ≤
=
i=0
+ ···
i
summary : if G0 (ξh) is unif power bdd, then so is G(ξh) = G0 (ξh) + kB
further theoretical results
von Neumann condition : max ρ(G(ξh)) ≤ 1 + kb
ξh
↑
spectral radius (largest magnitude of an e-value)
claim
1. The vN condition is necessary for stability in the 2-norm.
2. If p = 1, the vN condition is necessary and sufficient for stability in the 2-norm.
3. If p ≥ 2, the vN condition in not sufficient for stability in the 2-norm.
note
The Kreiss matrix theorem gives several conditions on G(ξh) which are necessary
and sufficient for stability in the 2-norm.
Lax equivalence theorem
Consider a linear constant coefficient PDE of the form vt = A0 v + A1 vx + A2 vxx
(for example) and a finite-difference scheme of the form un+1 = Qun . If the
initial value problem for the PDE is well-posed and the difference scheme is
consistent, then stability is necessary and sufficient for convergence.
82
convection-diffusion equation
vt + cvx = νvxx
, assume c > 0
un+1
− unj
j
+ cD− unj = νD+ D− unj
k
un+1
= unj − kcD− unj + νkD+ D− unj
j
= unj − cλ1 (unj − unj−1 ) + νλ2 (unj+1 − 2unj + unj−1 )
k
k
, λ2 = 2
h
h
consider stability in the ∞-norm
λ1 =
un+1
= (1 − cλ1 − 2νλ2 )unj + (cλ1 + νλ2 )unj−1 + νλ2 unj+1
j
then ||un+1 ||∞ ≤ ||un ||∞ ⇔ 1 − cλ1 − 2νλ2 ≥ 0
pf . . .
νλ2
1/2
cλ1
1
⇒















cλ1 ≤ 1 − 2νλ2 : more restrictive than for vt + cvx = 0
νλ2 ≤
1
2
− 12 cλ1 : · · · · · · · · · · · ” · · · · · · · · · · vt = νvxx
problem : if ν << 1, then artificial viscosity can overwhelm physical viscosity
analysis of numerical wave speed for leap-frog scheme
vt + cvx = 0
un+1
− un−1
j
j
+ cD0 unj = 0 , needs u0j , u1j to start
2k
83
!
un+1
j
unj
"
!
!
u? n+1 (ξh)
−2icλ sin ξh 1
=
u? n (ξh)
1
0
−2ckD0
=
1
"
1
0
!
"!
unj
un−1
j
"
"!
u? n (ξh)
u? n−1 (ξh)
e-values of G(ξh) : µ = −icλ sin ξh ±
"
@
1 − c2 λ2 sin2 ξh
We showed that G(ξh) is unif power bdd ⇔ |c|λ < 1. However even when the
scheme is stable, there is still a problem due to the extraneous root.
recall : numerical wave speed = c# =
µ1 = −icλ sin ξh +
@
1 − c2 λ2 sin2 ξh
θ(ξh)
, θ(ξh) = arg ρ(ξh)
−λξh
= −icλ(ξh − 16 (ξh)3 + · · ·) + 1 − 12 c2 λ2 (ξh)2 + · · ·
−cλξh + 16 cλ(ξh)3 + · · ·
tan θ1 =
1 − 12 c2 λ2 (ξh)2 + · · ·
:
same as LW
c#1 = c(1 − 16 (1 − c2 λ2 )(ξh)2 + · · ·)
µ2 = −icλ sin ξh −
@
1 − c2 λ2 sin2 ξh
= −icλ(ξh − 16 (ξh)3 + · · ·) − (1 − 12 c2 λ2 (ξh)2 + · · ·)
−cλξh + 16 cλ(ξh)3 + · · ·
tan θ2 =
−1 + 12 c2 λ2 (ξh)2 + · · ·
= (−cλξh + 16 cλ(ξh)3 + · · ·)(−1 − 12 c2 λ2 (ξh)2 + · · ·)
c#2 = c(−1 + 16 (1 − c2 λ2 )(ξh)2 + · · ·) :
wrong direction
recall : the numerical solution has components of the form unj = µn eijξh
µ1 = 1 − icλξh − 12 c2 λ2 (ξh)2 + · · · = e−icλξh + · · ·
unj = µn1 eijξh ≈ eiξ(xj −#c1 tn ) : genuine solution of PDE
µ2 = (−1)(1 + icλξh − 12 c2 λ2 (ξh)2 + · · ·) = −eicλξh + · · ·
unj = µn2 eijξh ≈ (−1)n eiξ(xj −#c2 tn ) : spurious solution
The spurious solution oscillates in time and travels in the wrong direction.