Harmonic functions.

Definition 3.4.1   Let $\phi$ be a function of two real variables $x$ and $y$ , 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:

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

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

$$\phi _x = \frac {y^2-x^2}{(x^2+y^2)^2};$$

$$\phi _y = \frac {-2xy}{(x^2+y^2)^2};$$

$$\phi _{xx} = \frac {2x(x^2-3y^2}{(x^2+y^2)^3};$$

$$\phi _{yy} = \frac {2x}{(3y^2-x^2)^3}.$$

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 $f=u+iv$ is analytic in a neighborhood $\mathcal{V}$ of $z_0$ . Moreover suppose that the partial derivatives of $u$ and $v$ are differentiable and that the second partial derivatives are continuous functions on $\mathcal{V}$ . From Cauchy-Riemann equations follows:

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

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

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

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

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

Example 3.4.4   Let $f(z)=z^3$ . The function is an entire function, as we proved previously. With $z=x+iy$ and $x,y \in \mathbb{R}$ , we have:

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

On the one hand, e have:

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

On the other hand, we have:

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

The functions $u$ and $v$ are both harmonic.

Definition 3.4.5   Let $u$ be a function of two real variables, harmonic over a domain $\mathcal{D}$ . Let $v$ be a function of two real variables, defined over $\mathcal{D}$ , and such that $f=u+iv$ is analytic over $\mathcal{D}$ . Then $v$ is called an harmonic conjugate of $u$ .

Example 3.4.6   Let $u(x,y)=x^2-y^2$ . This is a polynomial function, thus it has partial derivatives of any order.

(i)

The function $u$ 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 $v$ such that $f=u+iv$ is analytic over $\mathbb{C}$ , then $u$ and $v$ verify the Cauchy-Riemann equations (v.s. section 3):

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

where $C_1$ and $C_2$ are independent of $y$ and $x$ respectively. As these two formulas define the same function, $C_1$ and $C_2$ must be constant, i.e. $v(x,y)=2xy+k$ , where $k \in \mathbb{R}$ .

(iii)

Finally, we have:

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

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 $f$ is analytic over some domain $D$ . Then the level curves of $u$ are orthogonal to the level curves of $v$ .

Proof. Use Cauchy-Riemann equations and show that the gradients of $u$ and $v$ 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 $u(x,y)=x^2-y^2$ and $v(x,y)=2xy$ .

For general $k$ , then equation $u(x,y)=k$ defines an equilateral hyperbola, whose symmetry axes are the coordinate axes (red curves in Figure 2), and the equation $v(x,y)=k$ defines an equilateral hyperbola whose asymptotes are the coordinates axes (blue curves in Figure 2). For $k=0$ , 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

Proof. $\qedsymbol$