1. ## [SOLVED] Harmonic conjugate

Hi! I have this function:

$\displaystyle u(x,y)=y^{3}-3x^{2}y$

And I want it harmonic conjugate f(z) where:

$\displaystyle f(z)=u(z)+iv(z)$

so I use cauchy-riemann equations:

$\displaystyle u_{x}=v_{y} and u_{y}=-v_{x}$

So:

$\displaystyle u_{x}=-6yx and u_{y}=3y^{2}-3x^{2}$

$\displaystyle \int -6yx dy= -3xy^{2}+C$

$\displaystyle \int -(3y^{2}-3x^{2}) dx= -3y^{2}x+x^{3}+C$

Why have i two diferents values for $\displaystyle v(z)$?

Thanks!

2. $\displaystyle u_{x}(x,y) = -6xy$

$\displaystyle u_{y}(x,y) = 3y^{2} -3x^{2}$

$\displaystyle v(x,y) = \int u_{x} (x,y)dy$

$\displaystyle v(x,y) = -6 \int xy \ dy = -3xy^{2} + C(x)$

$\displaystyle v_{x}(x,y) = -3y^{2} + C'(x) = -u_{y}(x,y) = -3y^{2}+3x^{2}$

so $\displaystyle C'(x) = 3x^{2}$

then $\displaystyle C(x) = x^{3}$

and $\displaystyle v(x,y) = -3xy^{2} + x^{3} + C$

3. Originally Posted by Bop
Hi! I have this function:
Why have i two diferents values for $\displaystyle v(z)$?
Thanks!
Because when you partially integrate with respect to $\displaystyle x$ you forgot to consider an arbitrary function of $\displaystyle y$ that was lost (since any function of $\displaystyle y$ differentiated w.r.t. $\displaystyle x$ is 0) and vice versa.

Another way of solving your problem:

$\displaystyle u_x = -6xy, \quad u_y=3y^2-3x^2$

$\displaystyle v_y = -6xy$
$\displaystyle v_x = -u_y = 3x^2-3x^2$

$\displaystyle v = \int -6xy\,dy=-3y^2x+g(x)+C$ (integrate w.r.t. y)
$\displaystyle v = \int(3x^2-3y^2)\,dx=x^3-3y^2x+h(y)+C$ (integrate w.r.t. x)

Therefore, by inspection, $\displaystyle g(x)$ must be $\displaystyle x^3$, and $\displaystyle h(y) = 0$.

So $\displaystyle v = -3y^2x + x^3 + C$