Harmonic Conjugates of Analytic Complex Functions

Harmonic Conjugates of Analytic Complex Functions

Recall from the Harmonicity of the Real and Imaginary Parts of an Analytic Complex Function page that if $A \subseteq \mathbb{C}$ is open and $f : A \to \mathbb{C}$ is a function with $f = u + iv$ then if $f$ is analytic on $A$ we have that both $u$ and $v$ are harmonic on $A$, that is, the Laplacians of $u$ and $v$ are equal to zero on all of $A$, i.e.:

\begin{align} \quad \Delta (u) = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \quad \mathrm{and} \quad \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 0 \end{align}

We are now able to define special pairs of functions $u$ and $v$ for which $f = u + iv$ is analytic on some open subset $A$ of $\mathbb{C}$.

Definition: Let $A \subseteq \mathbb{C}$ be open and let $u, v : A \to \mathbb{R}$. If $f = u + iv$ is analytic on $A$ then $u$ and $v$ are called Harmonic Conjugates on $A$.

For example, suppose that we want to find a harmonic conjugate of $u(x, y) = x^2 - y^2$, call it $v = v(x, y)$. If $f = u + iv$ is to be analytic on $A$ then we know by the Cauchy-Riemann theorem that the Cauchy-Riemann equations are to be satisfied, that is:

\begin{align} \quad \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \mathrm{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \end{align}

The partial derivatives of $u$ are:

\begin{align} \quad \frac{\partial u}{\partial x} = 2x \quad \mathrm{and} \quad \frac{\partial u}{\partial y} = -2y \end{align}

With the first equation, $\displaystyle{\frac{\partial u}{\partial x} = 2x}$ we integrate with respect to $y$ to get:

\begin{align} \quad \int \frac{\partial u}{\partial x} \: dy = \int \frac{\partial v}{\partial y} \: dy \\ \quad \int 2x \: dy = v(x, y) \\ \quad 2xy + h(x) = v(x, y) \end{align}

Here, $h$ is a function of the variable $x$ which appears since differentiating the lefthand side with respect to $y$ eliminates $h(x)$. Now we take the derivative of both sides with respect to $x$ to get:

\begin{align} \quad \frac{\partial}{\partial x} [2xy + h(x)] = \frac{\partial v}{\partial x} \\ \quad 2y + h'(x) = \frac{\partial v}{\partial x} \end{align}

By the second Cauchy-Riemann equation we have that $-\frac{\partial u}{\partial y} = \frac{\partial v}{\partial x}$, that is, $2y = \frac{\partial v}{\partial x}$. This implies that $h'(x) = 0$ and so $h(x) = C$ for some constant $C$. Since $v(x, y) = 2xy + h(x)$ we have that the harmonic conjugates $v$ of $u$ are:

\begin{align} \quad v(x, y) = 2xy + C \end{align}

It is important to note that the harmonic conjugate of a function $u$ is not unique and may differ up to a constant $C$. In the example above if we choose $C = 0$, then $v(x, y) = 0$ and $f$ is given by:

\begin{align} \quad f(z) = f(x, y) = [x^2 - y^2] + 2xyi \end{align}

Notice that this simply means that $f(z) = z^2$ for each $z = x + yi \in \mathbb{C}$ which we know to be analytic on all of $\mathbb{C}$.

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