Proving the Harmonic Property of Analytic Functions

[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.

 

[Meir Achuz]

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

 

[ultimateguy]

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

 

[ultimateguy]

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.

 

[HallsofIvy]

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.

 

[ultimateguy]

Thanks for your reply.

There is a part b) to the problem, and it is this:

b) Show that

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

I solved it easily using the Cauchy-Riemann equations, so I figured that the hint was for the first part.