## 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.