# Definition of Harmonic Conjugates

This definition of Harmonic Conjugates is taken from Complex Variables and Applications by James Ward Brown, Ruel V. Churchill That is due to the above text $u,v:D(\subset\mathbb R^2)\to\mathbb R,D$ being a domain, are said to be the harmonic conjugates of each other if (1) $u,v$ have continuous partials of 1st and 2nd order in $D,$ (2) $u,v$ satisfy the Laplace equations in $D,$ (3) $u,v$ satisfy the C-R equations in $D.$

But I think condition (2) is an overstatement. Here's my logic. Please comment on my thoughts:

Due to the sufficient condition of differentiability, (1) and (3) implies the analycity of the function $u+iv$ in $D.$ As a result $u,v$ must satisfy the Laplace equations in $D.$

You are correct. In fact, if $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\tag{1}$$ and $$\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}\tag{2}$$ then taking the partial of $(1)$ with respect to $x$, we get $$\frac{\partial^2u}{\partial x^2}=\frac{\partial^2v}{\partial x\partial y}\tag{3}$$ and taking the partial of $(2)$ with respect to $y$, we get $$\frac{\partial^2v}{\partial y\partial x}=-\frac{\partial^2u}{\partial y^2}\tag{4}$$ Subtracting $(4)$ from $(3)$ and using the equality of mixed partials yields $$\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}=0\tag{5}$$ Similarly for $v$, by taking the partial of $(1)$ with respect to $y$ and adding the partial of $(2)$ with respect to $x$ yields. $$\frac{\partial^2v}{\partial x^2}+\frac{\partial^2v}{\partial y^2}=0\tag{6}$$