Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 10 pps
Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 10 pps
... (kab)
2
1 955
45. 5 Exercises
Exercise 45. 1
Consider the Cauchy problem for the diffusion equation with a source.
u
t
− κu
xx
= s(x, t), u(x, 0) = f(x), u → 0 as x → ±∞
Find the Green function for this ... =
c
1
(ω)
e
ωy
+c
2
(ω)
e
−ωy
, for ω = 0,
c
1
(ω) + c
2
(ω)y, for ω = 0.
Note that
e
ωy
is the bounded solution for ω < 0, 1 is the bounded solution for ω = 0 and
e
−...
Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 6 pps
... −
Y
Y
= −λ
We have differential equations for X and Y .
X
+ λX = 0, X(0) = X(1) = 0
Y
− λY = 0, Y (0) = 0
The eigenvalues and orthonormal eigenfunctions for X are
λ
n
= (nπ)
2
, X
n
(x) =
√
2 ... differential equation, we will expand the solution in a series of eigenfunctions in
x for which the coefficients are functions of t. The solution for u has the form,
u(x, t) =
∞
n=...
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 6 pps
... compute the volume of the intersecting cylinders.
-1
-0 .5
0
0 .5
1
-1
-0 .5
0
0 .5
1
-1
-0 .5
0
0 .5
1
-1
-0 .5
0
0 .5
1
-1
-0 .5
0
0 .5
1
Figure 5. 13: The intersection of the two cylinders.
174
Im(z)
Re(z)
r
(x,y)
θ
Figure ... (1
2
)
1/2
=
1
1/2
= ±1 and
1
1/2
2
= (±1)
2
= 1.
Example 6.6.2 Consider 2
1 /5
, (1 + ı)
1/3
and (2 + ı)
5/ 6
.
2
1 /5
=
5
√
2
e
ı2πk /5
,...
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 10 doc
... -1
-0 .5 0 .5
1
0 .5
1
1 .5
2
2 .5
3
Figure 7 .56 : The principal branch of the arc cosine, Arccos(x).
Solution 7.33
Arccos(x) is shown in Figure 7 .56 for real variables in the range ... log
(−1)
1/2
= 0.
350
0
z → z
1/2
z → z + 5
For the positive branch of g(z), the branch cut is mapped to the line x = 5 and the z plane is mapped to the
half-plane x > 5. log(w) ha...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 1 pps
... imaginary part of some analytic function.
Solution 8.11
We write the real and imaginary parts of f(z) = u + ıv.
u =
x
4/3
y
5/ 3
x
2
+y
2
for z = 0,
0 for z = 0.
, v =
x
5/ 3
y
4/3
x
2
+y
2
for ... equations for µ and ν are satisfied if and only if the Cauchy-Riemann
equations for u and v are satisfied. The continuity of the first partial derivatives of u and v impli...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 4 ppsx
... circle |z| = 1. On this circle:
|10| = 10
|z
6
− 5z
2
| ≤ |z
6
| + | −5z
2
| = 6
Since |z
6
− 5z
2
| < |10| on |z| = 1, p(z) has the same numb er of roots as 10 in |z| < 1. p(z) has no roots ... 11.4
Consider the circle |z| = 2. On this circle:
|z
6
| = 64
| −5z
2
+ 10| ≤ | − 5z
2
| + |10| = 30
Since |z
6
| < |−5z
2
+ 10| on |z| = 2, p(z) has the same number of roots as z
6
in...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 10 ppt
... numerator and denominator vanish, we apply L’Hospital’s rule.
= ı2π
2
k=0
lim
z→
e
ıπ(1+2k)/6
1
6z
5
=
ıπ
3
2
k=0
e
−ı 5( 1+2k)/6
=
ıπ
3
e
−ı 5/ 6
+
e
−ıπ 15/ 6
+
e
−ıπ 25/ 6
=
ıπ
3
e
−ı 5/ 6
+
e
−ıπ/2
+
e
−ıπ/6
=
ıπ
3
−
√
3 ... that
G(c)
=
πc
√
(c
2
−1)
3
for c > 1,
= −
πc
√
(c
2
−1)
3
for c < 1,
is divergent for −1 ≤ c ≤ 1.
In terms of F (a, b), t...
Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 6 ppsx
... integrating factor and integrate to obtain the solution.
d
dt
t
−4
u
= −2t
−6
u =
2
5
t
−1
+ ct
4
y
−2
=
2
5
t
−1
+ ct
4
y = ±
1
2
5
t
−1
+ ct
4
y = ±
√
5t
√
2 + ct
5
(b)
dy
dx
+ 2xy + ... =
e
R
−4t
−1
dt
=
e
−4 log t
= t
−4
.
We multiply by the integrating factor and integrate.
d
dt
t
−4
u
= −2t
−6
t
−4
u =
2
5
t
5
+ c
u =
2
5
t
−1
+ ct
4
101 1
18.6 *Equidimensio...
Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 7 pps
... Equations
101 7
19.6 Hints
The Constant Coefficient Equation
Normal Form
Hint 19.1
Transform the equation to normal form.
Transformations of the Independent Variable
Integral Equations
Hint 19.2
Transform ... that satisfy the left and right boundary conditions are
c
1
(x − a) and c
2
(x − b).
Thus the Green’s function has the form
G(x|ξ) =
c
1
(x − a), for x ≤ ξ
c
2
(x − b), for x ≥...
Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 8 ppsx
... > 0, α = 1.
107 2
0 .5
1
-0.3
-0.2
-0.1
0.1
0 .5
1
-0.3
-0.2
-0.1
0.1
0 .5
1
-0.3
-0.2
-0.1
0.1
0 .5
1
-0.3
-0.2
-0.1
0.1
Figure 21.3: Plot of G(x|0. 05) ,G(x|0. 25) ,G(x|0 .5) and G(x|0. 75) .
Thus the ... 1, . . . , n,
and W [y
1
, y
2
, . . . , y
n
](x) is the Wronskian of {y
1
(x), . . . , y
n
(x)}.
107 0
-4
-2 2
4
0. 05
0.1
0. 15
0.2
0. 25
0.3
-4
-2 2
4
-0.3
-0. 25
-0.2...
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