The divergence of holomorphic vector field

Question

Consider the holomorphic vector field. Why the divergence of this field is zero? We can write $f(z)=u(x,y)+iv(x,y)$ and the divergence is not equal to $\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}$? Or if this field are holomorphic, then we should consider only the fields of type $(u,-v)$?

Answer 1

If $f(z) = u + iv$ is holomorphic, it satisfies the Cauchy-Riemann equations;

$u_x = v_y, \tag 1$

$u_y = - v_x; \tag 2$

thus the divergence of the $2$-dimensional vector field $(u, v)$ is

$\nabla \cdot (u, v) = u_x + v_y = u_x + u_x = 2u_x \ne 0 \tag 3$

in general. For example, taking

$f(z) = z^2 = (x + iy)^2 = x^2 - y^2 + 2ixy, \tag 4$

we have

$u = x^2 - y^2, \tag 5$

$v = 2xy, \tag 6$

$u_x = 2x, \tag 7$

$v_y = 2x, \tag 8$

$\nabla \cdot (u, v) = u_x + v_y = 2x + 2x = 4x \ne 0. \tag 9$

We thus see that vector fields $(u, v)$ associated to holomorphic functions $f(z) = u(x, y) + i v(x, y)$ are not in general divergence free; indeed if

$u_x + v_y = \nabla \cdot (u, v) = 0, \tag{10}$

then

$u_x = -v_y, \tag{11}$

but by (1) we obtain

$u_x = - u_x \Longrightarrow u_x = 0 = v_y; \tag{12}$

it follows then that $u$ is a function of $y$ alone, and $v$ is a function of $x$ alone; the same is then true or $u_y$ and $v_x$; hence by (2), we must have $u_y$ and $v_x$ constant:

$u_y = - v_x = c \in \Bbb R; \tag{13}$

it then follows that $u$ and $v$ must take the form(s)

$u = cy + u_0, \; v = -cx + v_0; \tag{14}$

hence

$f(z) = u + iv = cy + u_0 - icx + iv_0 = c(y - ix) + (u_0 + iv_0)$ $= -ic(x + iy) + u_0 + iv_0 = -icz + \beta, \tag{15}$

where $\beta \in \Bbb C$ is an arbitrary complex constant. Holomorphic functions of the form (15) are the only possible ones satisfying (10).

On the other hand, if $f(z) = u + iv$ is holomorphic, then the vector field $(u, -v)$ is in fact divergence free:

$\nabla \cdot (u, -v) = u_x + (-v_y) = u_x - v_y = 0 \tag{16}$

by (1).

It is perhaps worth noting that similar concerns apply to the curl operator $\nabla \times$; we have

$\nabla \times (u, v) = v_x - u_y = -u_y - u_y = -2 u_y \ne 0; \tag{17}$

indeed, in the case $f(z) = z^2$,

$\nabla \times (x^2 - y^2, 2xy) = 2y - (-2y) = 4y \ne 0; \tag{18}$

on the other hand

$\nabla \times (u, - v) = -v_x - u_y = -v_x - (-v_x) = -v_x + v_x = 0. \tag{19}$

If $(u, v)$ is curl free,

$v_x - u_y = \nabla \times (u, v) = 0, \tag{20}$

so

$v_x = u_y = -v_x \tag{21}$

by (2); thus

$v_x = 0 = u_y; \tag{22}$

in this case $v$ depends only on $y$ and $u$ only on $x$; now by (1),

$u_x = v_y = c, \tag{23}$

$c$ constant, and

$u = cx + u_0, \tag{24}$

$c = cy + c_0, \tag{25}$

$u + iv = cx + u_0 + i(cy + v_0) = c(x + iy) + (u_0 + iv_0) = cz + \beta, \tag{26}$

$\beta \in \Bbb C$ again an arbitrary complex constant.