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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