• Support PF! Buy your school textbooks, materials and every day products Here!

Cauchy-Riemann conditions

  • #1
125
1

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

[tex]\bigtriangledown^2 u=\bigtriangledown^2 v=0[/tex]

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


Homework Equations


Cauchy-Riemann conditions:

[tex]\frac{\delta u}{\delta x} = \frac{\delta v}{\delta y}[/tex]
[tex]\frac{\delta u}{\delta y} = -\frac{\delta v}{\delta x}[/tex]


The Attempt at a Solution


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

[tex]\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[/tex]

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

Answers and Replies

  • #2
Meir Achuz
Science Advisor
Homework Helper
Gold Member
2,171
65
Your eqs. for del^2 are wrong.
[tex]\nabla^2 u=\partial_x\partial_x u+\partial_y\partial_y u.[/tex]
 
  • #3
125
1
Dang, you're right. Can I dot it into an element of length like this?

[tex]\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[/tex]
 
  • #4
125
1
There is a hint in the problem that says I need to construct vectors normal to the curves [tex]u(x,y)=c_i[/tex] and [tex]v(x,y)=c_j[/tex]. Wow, I'm pretty lost.
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,808
933
The Cauchy-Riemann equations are
[tex]\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}[/tex]
[tex]\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}[/tex]
which is what you have, allowing for your peculiar use of [itex]\delta[/itex] rather than [itex]\partial[/itex]!

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
[tex]\frac{\partial u}{\partial x}\vec{i}+ \frac{\partial u}{\partial y}\vec{j}[/tex]
and a normal vector to v(x,y)= c is
[tex]\frac{\partial v}{\partial x}\vec{i}+ \frac{\partial v}{\partial y}\vec{j}[/tex].
Using the Cauchy-Riemann equations, that second equation is
[tex]-\frac{\partial u}{\partial y}\vec{i}+ \frac{\partial u}{\partial x}\vec{j}[/tex]
which tells us the the two families of curves are orthogonal but that does not directly tell us about [itex]\nabla^2 u[/itex] and [itex]\nabla^2 v[/itex].
 
  • #6
125
1
Thanks for your reply.

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

b) Show that

[tex]\frac{\partial u}{\partial x}\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}\frac{\partial v}{\partial y} = 0[/tex]

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

Related Threads on Cauchy-Riemann conditions

  • Last Post
Replies
1
Views
651
  • Last Post
2
Replies
30
Views
3K
  • Last Post
Replies
4
Views
711
  • Last Post
Replies
16
Views
4K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
271
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
8
Views
989
  • Last Post
2
Replies
34
Views
3K
  • Last Post
Replies
2
Views
1K
Top