Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 6 pps
Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 6 pps
... +
4
3
P
2
(cos θ) exp
−
6a
2
R
2
t
Ψ(θ, t) = −
1
3
+
2 cos
2
θ −
2
3
exp
−
6a
2
R
2
t
Solution 37.34
Since we have homogeneous boundary conditions at x = 0 and x = 1, we will expand the solution ... −
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...
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 6 pps
... ellipse.
187
6. 8 Hints
Complex Numbers
Hint 6. 1
Hint 6. 2
Hint 6. 3
Hint 6. 4
Hint 6 .5
Hint 6. 6
Hint 6. 7
The Complex Plane
Hint 6. 8
Hint 6. 9
208
Part III
Functions of a Complex Variable
179
6. 7 Exercises
Complex ... (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 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 + ... 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
1011
18 .6 *Equidimensional-in-y Equations
A...
Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 6 pps
... no value of a for which both cos a and sin a vanish, the system is not orthogonal for any interval
of length π.
2. First note that
π
0
cos nx dx = 0 for n ∈ N.
If n = m, n ≥ 1 and m ≥ 0 then
π
0
cos ... 2m)x)
=
1
2
n−1
n
k=1
odd k
n
(n − k)/2
cos(kx).
1387
-3
-2
-1 1
2
3
2
4
6
8
10
-3
-2
-1 1
2
3
-10
-5
5
10
Figure 28.10: The Fourier Cosine and Sine Series of f(x) =...
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 ... → 0 as x → ∞,
and subject to the initial condition
G(x, τ
−
) = 0.
1. Solve for G with the Fourier cosine transform.
1 964
Taking the inverse Fourier sine transform of ˆu(ω, y)...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 1 pps
... 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 implies the same of
µ and ν. Thus ... =
x
3
(1+ı)−y
3
(1−ı)
x
2
+y
2
for z = 0,
0 for z = 0.
Show that the partial derivatives of u and v with respect to x and y exist at z = 0 and that u
x
= v
y
and u
y...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 4 ppsx
... 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 |z| < 2. p(z) has 6 roots in |z| < ... in |z| < 2.
Consider the 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 o...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 6 doc
... series converges absolutely for |z| < 1.
2.
∞
k=1
k
k
z
k
59 8
The series converges absolutely for |z| <
e
.
4.
∞
k=0
(z + 5)
2k
(k + 1)
2
We u se the ratio formula to determine the domain ... convergence.
lim
k→∞
(z + 5)
2(k+1)
(k + 2)
2
(z + 5)
2k
(k + 1)
2
< 1
|z + 5|
2
lim
k→∞
(k + 2)
2
(k + 1)
2
< 1
|z + 5|
2
lim
k→∞
2(k + 2)
2(k...
Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 7 pps
... Equations
1017
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 the ... 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...
Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 8 ppsx
... 0, α = 1.
1072
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 Green ... 1, . . . , n,
and W [y
1
, y
2
, . . . , y
n
](x) is the Wronskian of {y
1
(x), . . . , y
n
(x)}.
1070
-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
-0...
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