This definition of Harmonic Conjugates is taken from Complex Variables and Applications by James Ward Brown, Ruel V. Churchill

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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. $$ \frac{\partial^2v}{\partial x^2}+\frac{\partial^2v}{\partial y^2}=0\tag{6} $$

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  • $\begingroup$ So actually continuity along with C-R equations result Laplace equation? $\endgroup$ – user146075 Apr 28 '14 at 7:44
  • $\begingroup$ @user146075: continuity of the first and second order partials and the Cauchy-Riemann equations, yes; that is, conditions (1) and (3). $\endgroup$ – robjohn Apr 28 '14 at 7:53

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