# Harmonic conjugate function

• Jan 16th 2010, 12:17 AM
nice rose
Harmonic conjugate function
u= ln(x2+y2) is a harmonic function

find a harmonic conjugate function v and the analytic function f(z) ?

• Jan 16th 2010, 01:00 AM
shawsend
Try and be more precise with your notation. That's squares right? So first look at this thread:

http://www.mathhelpforum.com/math-he...xvariable.html

Then see if you can understand what I'm doing (lazy-style) with the following Mathematica code:

Code:

In[43]:= u[x_, y_] = Log[x^2 + y^2]; v[x_, y_] = FullSimplify[   Integrate[D[u[x, y], x], y] -     Integrate[D[Integrate[D[u[x, y], x],         y], x] + D[u[x, y], y], x]] FullSimplify[D[u[x, y], x] -   D[v[x, y], y]] FullSimplify[D[v[x, y], x] +   D[u[x, y], y]] Out[44]= 2*ArcTan[y/x] Out[45]= 0 Out[46]= 0
• Jan 16th 2010, 04:34 AM
HallsofIvy
Quote:

Originally Posted by nice rose
u= ln(x2+y2) is a harmonic function

find a harmonic conjugate function v and the analytic function f(z) ?

Any analytic function u(x,y)+ v(x,y)i must satisfy the "Cauchy-Riemann" equations: $\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}$ and [tex]\frac{\partial v}{\partial x}= -\frac{\partial u}{\partial y}.

The real and imaginary parts of any analytic function are harmonic functions and "v" is called the "harmonic conjugate" of u and vice versa.

Since $\frac{\partial ln(x^2+ y^2}{\partial x}= \frac{2x}{x^2+ y^2}$ and $\frac{\partial ln(x^2+ y^2}{\partial y}= \frac{2y}{x^2+ y^2}$, the Cauchy-Riemann equations become $\frac{\partial v}{\partial y}= \frac{2x}{x^2+ y^2}$ and $\frac{\partial v}{\partial x}= -\frac{2y}{x^2+ y^2}$.

The integral $\int \frac{A}{B+ y^2}dy= \frac{A}{B}\int \frac{1}{1+ \left(\frac{y}{\sqrt{B}}\right)^2} dy$ so the substitution $u= \frac{y}{\sqrt{B}}$ gives an "arctan" integral.