differential equations level 2
Partial Differential Equations part 2
Ngày tải lên :
07/11/2013, 19:15
... equation (19.1 .20 ) is
r
n+1
j+1 /2
− r
n
j+1 /2
∆t
=
s
n+1 /2
j+1
− s
n+1 /2
j
∆x
s
n+1 /2
j
− s
n−1 /2
j
∆t
= v
r
n
j+1 /2
− r
n
j −1 /2
∆x
(19.1.35)
If you substitute equation (19.1 .22 ) in equation ... propagation v
∂
2
u
∂t
2
= v
2
∂
2
u
∂x
2
(19.1 .2)
836
Chapter 19. Partial Differential Equations
Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -43108-5)
Copyright ... ≡
v∆t
∆x
(19.1.40)
Then
ξ =1−iα sin k∆x − α
2
(1 − cos k∆x)(19.1.41)
so
|ξ|
2
=1−α
2
(1 − α
2
)(1 − cos k∆x)
2
(19.1. 42)
The stability criterion |ξ|
2
≤ 1 is therefore α
2
≤ 1, or v∆t ≤ ∆x as usual.
Incidentally,...
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Tài liệu Integration of Ordinary Differential Equations part 2 pptx
Ngày tải lên :
15/12/2013, 04:15
... the whole interval. Figure 16.1 .2 illustrates the
idea. In equations,
k
1
= hf(x
n
,y
n
)
k
2
=hf
x
n
+
1
2
h, y
n
+
1
2
k
1
y
n+1
= y
n
+ k
2
+ O(h
3
)
(16.1 .2)
As indicated in the error term, ... organization about it:
k
1
= hf(x
n
,y
n
)
k
2
=hf(x
n
+
h
2
,y
n
+
k
1
2
)
k
3
=hf(x
n
+
h
2
,y
n
+
k
2
2
)
k
4
=hf(x
n
+ h, y
n
+ k
3
)
y
n+1
= y
n
+
k
1
6
+
k
2
3
+
k
3
3
+
k
4
6
+ O(h
5
)(16.1.3)
The ... Publications, New York),
§
25 .5. [1]
Gear, C.W. 1971,
Numerical Initial Value Problems in Ordinary Differential Equations
(Englewood
Cliffs, NJ: Prentice-Hall), Chapter 2. [2]
Shampine, L.F., and...
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Partial Differential Equations part 1
Ngày tải lên :
28/10/2013, 22:15
... velocity of propagation v
∂
2
u
∂t
2
= v
2
∂
2
u
∂x
2
(19.1 .2)
Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -43108-5)
Copyright (C) 1988-19 92 by Cambridge University ... equation
∂
2
u
∂t
2
= v
2
∂
2
u
∂x
2
(19.0.1)
where v = constant is the velocity of wave propagation. The prototypicalparabolic
equation is the diffusion equation
∂u
∂t
=
∂
∂x
D
∂u
∂x
(19.0 .2)
where ... representation (see
Figure 19.0 .2) ,
u
j+1,l
− 2u
j,l
+ u
j−1,l
∆
2
+
u
j,l+1
− 2u
j,l
+ u
j,l−1
∆
2
= ρ
j,l
(19.0.5)
or equivalently
u
j+1,l
+ u
j−1,l
+ u
j,l+1
+ u
j,l−1
− 4u
j,l
=∆
2
ρ
j,l
(19.0.6)
To...
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Free grammar ebook level 2
Ngày tải lên :
09/11/2013, 19:18
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• “I’ve lived here for 8 years.”
Since is used with a point in time, and means “from that point in time until
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“I’ve been married for over 20 years.”
“So have I.”
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Simple Past and Past Continuous
The past continuous is often...
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Tài liệu Integration of Ordinary Differential Equations part 1 doc
Ngày tải lên :
15/12/2013, 04:15
... the whole interval. Figure 16.1 .2 illustrates the
idea. In equations,
k
1
= hf(x
n
,y
n
)
k
2
=hf
x
n
+
1
2
h, y
n
+
1
2
k
1
y
n+1
= y
n
+ k
2
+ O(h
3
)
(16.1 .2)
As indicated in the error term, ... involving ordinary differential equations (ODEs) can always be
reduced to the study of sets of first-order differential equations. For example the
second-order equation
d
2
y
dx
2
+ q(x)
dy
dx
= ... generic problem in ordinary differential equations is thus reduced to the
study of a set of N coupled first-order differential equations for the functions
y
i
,i=1 ,2, ,N, having the general form
dy
i
(x)
dx
=...
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Tài liệu Partial Differential Equations part 3 pptx
Ngày tải lên :
15/12/2013, 04:15
... (19 .2. 19) write
D
j+1 /2
=
1
2
D(u
n
j+1
)+D(u
n
j
)
(19 .2. 22)
Implicit schemes are not as easy. The replacement (19 .2. 22) with n → n +1leaves
us with a nasty set of coupled nonlinear equations ... of (19 .2. 8) leads to
i
ψ
n+1
j
− ψ
n
j
∆t
= −
ψ
n+1
j+1
− 2
n+1
j
+ ψ
n+1
j −1
(∆x)
2
+ V
j
ψ
n+1
j
(19 .2. 27)
for which
ξ =
1
1+i
4∆t
(∆x)
2
sin
2
k∆x
2
+ V
j
∆t
(19 .2. 28)
This ... = D(u)du (19 .2. 23)
analytically for z(u), then the right-hand side of (19 .2. 1) becomes ∂
2
z/∂x
2
,which
we difference implicitly as
z
n+1
j+1
− 2z
n+1
j
+ z
n+1
j −1
(∆x)
2
(19 .2. 24)
Now linearize...
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Tài liệu Partial Differential Equations part 4 ppt
Ngày tải lên :
15/12/2013, 04:15
... ∆t /2.
In each substep, a different dimension is treated implicitly:
u
n+1 /2
j,l
= u
n
j,l
+
1
2
α
δ
2
x
u
n+1 /2
j,l
+ δ
2
y
u
n
j,l
u
n+1
j,l
= u
n+1 /2
j,l
+
1
2
α
δ
2
x
u
n+1 /2
j,l
+ δ
2
y
u
n+1
j,l
(19.3.16)
The ... second-order accurate
and unitary:
e
−iHt
1 −
1
2
iH∆t
1+
1
2
iH∆t
(19 .2. 35)
In other words,
1+
1
2
iH∆t
ψ
n+1
j
=
1 −
1
2
iH∆t
ψ
n
j
(19 .2. 36)
On replacing H by its finite-difference approximation ... L
1
piece; likewise U
2
, U
m
. Then a method of getting from
u
n
to u
n+1
is
u
n+1/m
= U
1
(u
n
, ∆t/m)
u
n +2/ m
= U
2
(u
n+1/m
, ∆t/m)
···
u
n+1
= U
m
(u
n+(m−1)/m
, ∆t/m)
(19.3 .22 )
The timestep for...
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Tài liệu Solution of Linear Algebraic Equations part 2 ppt
Ngày tải lên :
15/12/2013, 04:15
... equation
a
11
a
12
a
13
a
14
a
21
a
22
a
23
a
24
a
31
a
32
a
33
a
34
a
41
a
42
a
43
a
44
·
x
11
x
21
x
31
x
41
x
12
x
22
x
32
x
42
x
13
x
23
x
33
x
43
y
11
y
12
y
13
y
14
y
21
y
22
y
23
y
24
y
31
y
32
y
33
y
34
y
41
y
42
y
43
y
44
=
b
11
b
21
b
31
b
41
b
12
b
22
b
32
b
42
b
13
b
23
b
33
b
43
1000
0100
0010
0001
(2. 1.1)
Here ... equation
a
11
a
12
a
13
a
14
a
21
a
22
a
23
a
24
a
31
a
32
a
33
a
34
a
41
a
42
a
43
a
44
·
x
11
x
21
x
31
x
41
x
12
x
22
x
32
x
42
x
13
x
23
x
33
x
43
y
11
y
12
y
13
y
14
y
21
y
22
y
23
y
24
y
31
y
32
y
33
y
34
y
41
y
42
y
43
y
44
=
b
11
b
21
b
31
b
41
b
12
b
22
b
32
b
42
b
13
b
23
b
33
b
43
1000
0100
0010
0001
(2. 1.1)
Here ... vector):
a
11
a
12
a
13
a
14
0 a
22
a
23
a
24
00a
33
a
34
000a
44
·
x
1
x
2
x
3
x
4
=
b
1
b
2
b
3
b
4
(2. 2.1)
Here the primes signify...
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Tài liệu Partial Differential Equations part 5 ppt
Ngày tải lên :
15/12/2013, 04:15
... then adding the three equations,
we get
u
j 2
+ T
(1)
· u
j
+ u
j +2
= g
(1)
j
∆
2
(19.4. 32)
This is an equation of the same form as (19.4 .29 ), with
T
(1)
=21 −T
2
g
(1)
j
=∆
2
(g
j−1
−T·g
j
+g
j+1
)
(19.4.33)
After ... that λ
(r)
k
will involve terms like
cos (2 k/L) − 2 raised to a power. Solve the tridiagonal systems foru
k
j
at the levels
j =2
r
,2 2
r
,4 2
r
, , J − 2
r
. Fourier synthesize to get the y-values ... Similarly,
ρ
jl
=
1
JL
J−1
m=0
L−1
n=0
ρ
mn
e
2 ijm/J
e
2 iln/L
(19.4.3)
If we substitute expressions (19.4 .2) and (19.4.3) in our model problem (19.0.6),
we find
u
mn
e
2 im/J
+ e
2 im/J
+ e
2 in/L
+ e
2 in/L
− 4
=ρ
mn
∆
2
(19.4.4)
or
u
mn
=
ρ
mn
∆
2
2
cos
2 m
J
+cos
2 n
L
−...
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Tài liệu Partial Differential Equations part 6 doc
Ngày tải lên :
15/12/2013, 04:15
... turns out to be
ρ
s
1 −
π
2
2J
2
(19.5.11)
The number of iterations r required to reduce the error by a factor of 10
−p
is thus
r
2pJ
2
ln 10
π
2
1
2
pJ
2
(19.5. 12)
In other words, the number ... (19.5.31)
implicitly in two half-steps:
u
n+1 /2
− u
n
∆t /2
= −
L
x
u
n+1 /2
+ L
y
u
n
∆
2
− ρ
u
n+1
− u
n+1 /2
∆t /2
= −
L
x
u
n+1 /2
+ L
y
u
n+1
∆
2
− ρ
(19.5.35)
(cf. equation 19.3.16). Here we ... that λ
(r)
k
will involve terms like
cos (2 k/L) − 2 raised to a power. Solve the tridiagonal systems foru
k
j
at the levels
j =2
r
,2 2
r
,4 2
r
, , J − 2
r
. Fourier synthesize to get the y-values...
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Tài liệu Root Finding and Nonlinear Sets of Equations part 2 docx
Ngày tải lên :
15/12/2013, 04:15
... f1,f2;
if (*x1 == *x2) nrerror("Bad initial range in zbrac");
f1=(*func)(*x1);
f2=(*func)(*x2);
for (j=1;j<=NTRY;j++) {
if (f1*f2 < 0.0) return 1;
if (fabs(f1) < fabs(f2))
f1=(*func)(*x1 ... 3 52
Chapter 9. Root Finding and Nonlinear Sets of Equations
Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -43108-5)
Copyright (C) 1988-19 92 by Cambridge ... < 0.0) return 1;
if (fabs(f1) < fabs(f2))
f1=(*func)(*x1 += FACTOR*(*x1-*x2));
else
f2=(*func)(*x2 += FACTOR*(*x2-*x1));
}
return 0;
}
Alternatively, you might want to “look inward” on an initial...
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