, defined over a domain $\mathcal{D}$ . Suppose that $\phi$ has second order partial derivatives on $\mathcal{D}$ . The function $\phi$ is called an harmonic function if it verifies the equation:

$\displaystyle \phi_{xx}+\phi_{yy}=0$

Example 3.4.2 Take $\phi (x,y)= x/(x^2+y^2)$ . Then over $\mathbb{R}^2-{(0,0)}$ we have:

$\displaystyle \phi _x$	$\displaystyle = \frac {y^2-x^2}{(x^2+y^2)^2};$
$\displaystyle \phi _y$	$\displaystyle = \frac {-2xy}{(x^2+y^2)^2};$
$\displaystyle \phi _{xx}$	$\displaystyle = \frac {2x(x^2-3y^2}{(x^2+y^2)^3};$
$\displaystyle \phi _{yy}$	$\displaystyle = \frac {2x}{(3y^2-x^2)^3}.$

It follows that $\phi_{xx}+\phi_{yy}=0$ over $\mathbb{R}^2-{(0,0)}$ , i.e. $\phi$ is harmonic $\mathbb{R}^2-{(0,0)}$ .

Now suppose that is analytic in a neighborhood $\mathcal{V}$ of . Moreover suppose that the partial derivatives of and are differentiable and that the second partial derivatives are continuous functions on $\mathcal{V}$ . From Cauchy-Riemann equations follows:

$\begin{displaymath}\begin{cases}u_{xx}=v_{yx} \ u_{xy}=v_{yy} \end{cases} \qqua... ...\qquad \begin{cases}u_{yx}=-v{xx} \ u_{yy}=-v_{xy} \end{cases}\end{displaymath}$

As the second partial derivatives are continuous on $\mathcal{V}$ , we have $u_{xy}=u_{yx}$ and $v_{xy}+v_{yx}$ . It follows that:

$\displaystyle u_{xx}+u_{yy}=0$ and $\displaystyle \qquad v_{xx}+v_{yy}=0$

The computations that we performed before Definition 4.1 can be summarized in a theorem:

Theorem 3.4.3 If is a function of a complex variable, analytic over a domain $\mathcal{D} \subset \mathbb{C}$ , then and are harmonic over $\mathcal{D}$ .

Example 3.4.4 Let

. The function is an entire function, as we proved previously. With

and $x,y \in \mathbb{R}$ , we have:

$\displaystyle f(z)=(x+iy)^3=\underbrace{x^3-3xy^2)}_{=u(x,y)}+i \underbrace{(3x^2y-y^3)}_{=v(x,y)}.$

On the one hand, e have:

$\begin{displaymath}\begin{cases}u_x=3x^2-3y^2 \ u_y= -6xy \end{cases} \Longrigh... ...}=6x \ u_{yy}=-6x \end{cases} \Longrightarrow u_{xx}+u_{yy}=0.\end{displaymath}$

On the other hand, we have:

$\begin{displaymath}\begin{cases}v_x=6xy \ v_y= 3x^2-3y^2 \end{cases} \Longright... ...}=6y \ v_{yy}=-6y \end{cases} \Longrightarrow v_{xx}+v_{yy}=0.\end{displaymath}$

The functions

and

are both harmonic.

Definition 3.4.5 Let

be a function of two real variables, harmonic over a domain $\mathcal{D}$ . Let

be a function of two real variables, defined over $\mathcal{D}$ , and such that

is analytic over $\mathcal{D}$ . Then

is called an harmonic conjugate of

Example 3.4.6 Let

. This is a polynomial function, thus it has partial derivatives of any order.

(i)

The function

is harmonic:

$\begin{align*}\begin{cases}u_x=2x \ u_y=-2y \end{cases} \Longrightarrow \begin{... ...{xx}=2 \ u_{yy}=-2 \end{cases} \Longrightarrow u_{xx}+u_{yy}=2-2=0.\end{align*}$

(ii)

If there exists

such that

is analytic over $\mathbb{C}$ , then

and

verify the Cauchy-Riemann equations (v.s. section 3):

$\begin{displaymath}\begin{cases}v_x=-u_y \ v_y=u_x \end{cases} \Longrightarrow ... ...\begin{cases}v(x,y)=2xy+C_1(y) \ v(x,y)=2xy+C_2(x) \end{cases}\end{displaymath}$

where

and

are independent of

and

respectively. As these two formulas define the same function,

and

must be constant, i.e.

, where $k \in \mathbb{R}$ .

(iii)

Finally, we have:

$\displaystyle f(z)= x^2-y^2 +i(2xy+k).$

Please compare this with example 2.1.

We can now discover another important property of the analytic conjugates.

Theorem 3.4.7 Let $f(z)=u(x,y)+i\; v(x,y)$ , as usual. Suppose that is analytic over some domain . Then the level curves of are orthogonal to the level curves of .

Proof. Use Cauchy-Riemann equations and show that the gradients of

and

are orthogonal, whence the result. $\qedsymbol$

Example 3.4.8 Take $f(z)=z^2, \; z \in \mathbb{C}$ . With the usual algebraic notation, we have

and

For general , then equation defines an equilateral hyperbola, whose symmetry axes are the coordinate axes (red curves in Figure 2), and the equation defines an equilateral hyperbola whose asymptotes are the coordinates axes (blue curves in Figure 2). For , we have the union of the angle bisectors of the coordinate axes and the union of the coordinate axes respectively.

**Figure 2:** Level curves for two harmonic conjugates
$\begin{figure}\mbox{ \epsfig{file=OrthogonalLevelCurves,height=5cm} }\end{figure}$

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