I have a simple harmonic function $u(x,y) = x^2 – y^2 +2xy$ and wish to find its harmonic conjugate.
To find a harmonic conjugate $v$ of $u$, we must have
$$
u_x(x,y) =v_y(x,y)
$$
and
$$
u_y(x,y) = -v_x(x,y)
$$
From the first we have $v_y= u_x = 2x+2y \implies v = \int (2x+2y)dy = 2xy +y^2 + C(x)$. It now follows from the second equation that
$$
2(x-y) = -v_x =\implies -2(x-y) = v_x \implies v = 2xy-xy^2 + C(y)
$$
Therefore, we have
$$
2y +y^2 + C(x) = 2xy-xy^2 + C(y)
$$
How do I solve for $C(x)$ and $C(y)$?
Best Answer
Note that You started as
$$ v_y= u_x = 2x+2y \implies v = 2xy +y^2 +C(x) $$
To determine $C(x)$ we have
$$ v_x = 2y+0 + C'(x) $$
then use $v_x=-u_y$ to find $C(x)$ by solving the first order differential equation.