2(b) Given the function: u(x.y) = exp(-x) [x sin y – y cos
y]
Find v(x,y) such that f(z) = u + iv is analytic
Solution:
a) ¶ u/ ¶ x = exp(-x) (sin y) - x exp(-x) sin y + y (exp(-x) ) cos y = ¶ v/ ¶ y
b) - ¶ u/ ¶ y = - x exp (-x) cos y
- y (exp(-x) sin y +
exp(-x) cos y = ¶ v/ ¶ x
exp(-x) cos y = ¶ v/ ¶ x
Integrate
(a) with respect to y, keeping x constant so that:
v = - exp(-x) cos y + x exp (-x) cos y - exp(-x)[ y sin y + cos y) + F(x)
v = y
exp(-x) sin y + x exp(-x) cos y + F(x)
Here, F(x) is an arbitrary real function of x.
Substituting
the last result for v into the Cauchy equation for
¶ v/ ¶ x we get:
¶ v/ ¶ x we get:
y exp(-x) sin y – x exp(-x) cos y + exp (-x) cos y + F’(x)
= - y exp (-x) sin y
– x exp (-x) cos y – y exp(-x) sin y
Or: F’(x) = 0 and
F(x) = c (constant) Then, from the earlier
expression for v:
v = exp (-x) (y sin
y + x cos y) + c
3) Let f(z) = exp(x) cos(y) + i(exp(x)sin(y) = u(x,y) + iv(x,y)
a) Determine if the function is analytic for both u and v.
b) Determine if the function is harmonic for both u and v
a) Determine if the function is analytic for both u and v.
b) Determine if the function is harmonic for both u and v
Solutions:
a)
¶ u/ ¶ x
= exp (x) cos y
And: ¶
v/ ¶ y = exp (x) cos y
so: ¶ u/ ¶ x
= ¶ v/ ¶ y
SO the function is analytic for u.
Now, ¶ v/ ¶ x = exp(x) sin y and :
- ¶ u/ ¶ y = - exp
(x) [- sin y] = exp(x) sin y
SO: ¶ v/ ¶ x = - ¶ u/ ¶ y
Therefore, the function is also analytic for v.
b) ¶
2 u/ ¶
x2 = ¶ / ¶ x [ exp (x) cos y ] = exp (x) cos y
¶ 2 u/ ¶ y2 =
¶ / ¶ y
[- exp (x) sin y ] = - exp(x)
cos y
Then: ¶
2 u/ ¶
x2 + ¶ 2 u/ ¶ y2 = exp
(x) cos y +
(- exp(x) cos y) = 0
(- exp(x) cos y) = 0
So
the function is harmonic for u.
Looking
now at v:
¶ 2 v/ ¶ x2 =
¶ / ¶ x
[exp(x) sin
y ] =
exp(x) sin y
¶ 2 v/ ¶ y2 =
¶ / ¶ y [exp (x) cos y ] = exp
(x) [- sin (y)]
= - exp (x) sin y
= - exp (x) sin y
Then: ¶ 2 v/ ¶ x2 + ¶ 2 v/ ¶ y2
= exp(x) sin y + (- exp (x) sin y) = 0
= exp(x) sin y + (- exp (x) sin y) = 0
So
the function is also harmonic for v.
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