Differentiation of Functions of a Complex Variable 4: Harmonic Functions

Throughout this course, a connected open subset of $\mathbb{C}$ is called a domain. Suppose that a function $f(z)=u(x,y)+iv(x,y)$ is analytic in a domain $\mathcal{D}$. Then $f(z)$ satisfies the Cauchy-Riemann conditions. The Cauchy-Riemann conditions are also called the Cauchy-Riemann equations. Differentiating the Cauchy-Riemann equations with respect to $x$, we obtain
\begin{equation}
\label{eq:cr3}
u_{xx}=v_{yx},\ u_{yx}=-v_{xx}
\end{equation}
Differentiating the Cauchy-Riemann equations with respect to $y$, we also obtain
\begin{equation}
\label{eq:cr4}
u_{xy}=v_{yy},\ u_{yy}=-v_{xy}
\end{equation}
By the continuity of the partial derivatives of $u$ and $v$, we have
\begin{equation}
\label{eq:cr5}
u_{xy}=u_{yx},\ v_{xy}=v_{yx}
\end{equation}
Applying the last set of equations to each of the preceding two sets of equations, we obtain the Laplace equations
\begin{equation}
\label{eq:harmonic}
\begin{aligned}
\Delta u&=u_{xx}+u_{yy}=0\\
\Delta v&=v_{xx}+v_{yy}=0
\end{aligned}
\end{equation}
That is, $u$ and $v$ are harmonic  functions in $\mathcal{D}$.

Example. The function $f(z)=e^{-y}\sin x-ie^{-y}\cos x$ is entire, so both $u(x,y)=e^{-y}\sin x$ and $v(x,y)=-e^{-y}\cos x$ are harmonic in $\mathbb{C}$.

Definition. If two function $u$ and $v$ are harmonic in a domain $\mathcal{D}$ and their first-order partial derivatives satisfy the Cauchy-Riemann conditions throughout $\mathcal{D}$, $v$ is said to be a \emph{harmonic conjugate of $u$}.

Theorem. A function $f(z)=u(x,y)+iv(x,y)$ is analytic in a domain $\mathcal{D}$ if and only if $v$ is a harmonic conjugate of $u$.

Remark. If $v$ is a harmonic conjugate of $u$ in a domain $\mathcal{D}$, it is not necessarily true that $u$ is a harmonic conjugate of $v$ in $\mathcal{D}$ as seen in the following example.

Example. Let us consider $f(z)=z^2=x^2-y^2+i2xy$. Since $f(z)$ is entire, $v(x,y)=2xy$ is a harmonic conjugate of $u(x,y)=x^2-y^2$. However, $u$ cannot be a harmonic conjugate of $v$ since $2xy+i(x^2-y^2)$ is nowhere analytic (it is differentiable only at the origin $(0,0)$).

Example. [Finding a harmonic conjugate of a harmonic function] Let $u(x,y)=y^3-3x^2y$ and let $v(x,y)$ be a harmonic conjugate of $u(x,y)$. Then $u$ and $v$ satisfy the Cauchy-Riemann equations. It follows from the Cauchy-Riemann equation $u_x=v_y$ that $v_y=-6xy$. Integrating $v_y$ with respect to $y$, we obtain
$$v(x,y)=-3xy^2+\phi(x)$$
where $\phi(x)$ is some unknown of $x$. We determine $\phi(x)$ using $u_y=-v_x$: Differentiating $v(x,y)$ with respect to $x$, we have
$$v_x=-3y^2+\phi'(x)$$
Comparing this with
$$-u_y=-3y^2+3x^2$$
we get that $\phi'(x)=3x^2$ and so $\phi(x)=x^3+C$ where $C$ is a constant. Hence, we find a harmonic conjugate of $u(x,y)$
$$v(x,y)=-3xy^2+x^3+C$$
The corresponding analytic function is
$$f(z)=y^3-3x^2y+i(-3xy^2+x^3+C)$$