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:

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

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 $ 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{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 $ 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:

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

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 $ 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.
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*}    

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{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 $ 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}$ .
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 $ 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$

Noah Dana-Picard 2007-12-24