1. ## Harmonic conjugate function

u= ln(x2+y2) is a harmonic function

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

2. 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

3. 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.