Friday, October 11, 2013

Math Solutions (Part II) : Cauchy- Riemann Equations


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 

 
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:

 
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

 
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

 
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

 
 Then: 2 v/ x2      +  2 v/ y2   

=  exp(x) sin y +  (- exp (x) sin y) = 0

 
So the function is also harmonic for v.

 

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