# Cauchy-Riemann conditions

• ultimateguy

## Homework Statement

The functions u(x,y) and v(x,y) are the real and imaginary parts, respectively, of an analytic function w(z).
Assuming that the required derivatives exist, show that

$$\bigtriangledown^2 u=\bigtriangledown^2 v=0$$

Solutions of Laplace's equation such as u(x,y) and v(x,y) are called harmonic functions.

## Homework Equations

Cauchy-Riemann conditions:

$$\frac{\delta u}{\delta x} = \frac{\delta v}{\delta y}$$
$$\frac{\delta u}{\delta y} = -\frac{\delta v}{\delta x}$$

## The Attempt at a Solution

I expanded $$\bigtriangledown^2 u = \frac{\delta u}{\delta x}\frac{\delta u}{\delta x} + \frac{\delta u}{\delta y}\frac{\delta u}{\delta y}$$ and using the Cauchy-Riemann conditions I found

$$\bigtriangledown^2 u = \frac{\delta v}{\delta y}\frac{\delta v}{\delta y} + \frac{\delta v}{\delta x}\frac{\delta v}{\delta x}=\bigtriangledown^2 v$$

What I can't figure out how to do is prove that this is equal to zero.

Your eqs. for del^2 are wrong.
$$\nabla^2 u=\partial_x\partial_x u+\partial_y\partial_y u.$$

Dang, you're right. Can I dot it into an element of length like this?

$$\bigtriangledown^2 u \bullet d\vec{r}^2 = \frac{\delta}{\delta x}\frac{\delta u}{\delta x} dx^2 + \frac{\delta}{\delta y}\frac{\delta u}{\delta y} dy^2$$

There is a hint in the problem that says I need to construct vectors normal to the curves $$u(x,y)=c_i$$ and $$v(x,y)=c_j$$. Wow, I'm pretty lost.

The Cauchy-Riemann equations are
$$\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}$$
$$\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$
which is what you have, allowing for your peculiar use of $\delta$ rather than $\partial$!

Now just do the obvious: differentiate both sides of the first equation with respect to x and differentiate both sides of the second equation with respect to y and compare them.

Are you sure that the hint is for this particular problem? A normal vector to u(x,y)= c is
$$\frac{\partial u}{\partial x}\vec{i}+ \frac{\partial u}{\partial y}\vec{j}$$
and a normal vector to v(x,y)= c is
$$\frac{\partial v}{\partial x}\vec{i}+ \frac{\partial v}{\partial y}\vec{j}$$.
Using the Cauchy-Riemann equations, that second equation is
$$-\frac{\partial u}{\partial y}\vec{i}+ \frac{\partial u}{\partial x}\vec{j}$$
which tells us the the two families of curves are orthogonal but that does not directly tell us about $\nabla^2 u$ and $\nabla^2 v$.

$$\frac{\partial u}{\partial x}\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}\frac{\partial v}{\partial y} = 0$$