Analyticity and Laplacian Operator in Complex Functions: A Domain D Study

  • Thread starter Thread starter FanofAFan
  • Start date Start date
  • Tags Tags
    Domain
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
8 replies · 2K views
FanofAFan
Messages
43
Reaction score
0

Homework Statement


Let f(z) be analytic on a domain D. Let [tex]\Delta[/tex] = ([tex]\frac{\partial^2}{\partial x^2}[/tex] + [tex]\frac{\partial^2}{\partial y^2}[/tex]) Set W = |f(z)|2 show that W[tex]\Delta[/tex]W = (Wx)2+(Wy)2

Homework Equations


The Attempt at a Solution


W = (U2+V2)
[tex]\Delta[/tex]W = ([tex]\frac{\partial^2}{\partial x^2}[/tex] + [tex]\frac{\partial^2}{\partial y^2}[/tex]) (U2+V2)

also 4(U2+V2)[(Vx)2+(Uy)2]
 
Last edited:
Physics news on Phys.org
Well, firstly kindly clean up your texing. The latex code for superscript is more concisely - ^{...} .

As to your question. Do the following.

1. Write [tex]z = x + i y,\ \bar{z} = x- i y[/tex]. The rewrite [tex]\partial_x,\partial_y,\ \Delta[/tex] in terms of [itex]z[/itex] and [itex]\bar{z}[/itex]. (using chain rule for partial differentials.)

2. Note that [tex]W = |f(z)|^2 = f(z)\bar{f(z)} = f(z)f(\bar{z})[/tex]

Now solve.
 
Sorry about the latex coding or whatever... Thanks!
 
I'm confused... so would the [tex]\partial[/tex]x = (1 + i[tex]\partial[/tex]y/[tex]\partial[/tex]x)(1 - i[tex]\partial[/tex]y/[tex]\partial[/tex]x) is it just [tex]\textit{z}[/tex][tex]\overline{z}[/tex]?
 
Remember now, z and [tex]\bar{z}[/tex] are my independent variables.

[tex]\partial_x = \frac{\partial x}{\partial z}\partial_z+\frac{\partial x}{\partial \bar{z}}\partial_{\bar{z}} = \partial_z + \partial_\bar{z}[/tex]

Similarly, you can do it for y.
 
praharmitra said:
Remember now, z and [tex]\bar{z}[/tex] are my independent variables.

[tex]\partial_x = \frac{\partial x}{\partial z}\partial_z+\frac{\partial x}{\partial \bar{z}}\partial_{\bar{z}} = \partial_z + \partial_\bar{z}[/tex]

Similarly, you can do it for y.

so what's the difference between [tex]\partial_x[/tex] and other one with x?
 
FanofAFan said:
so what's the difference between [tex]\partial_x[/tex] and [tex]\partial[/tex]x?

Oh, I'm sorry if I didn't clarify my notation. It is standard to call

[tex]\frac{\partial}{\partial x}[/tex] as [tex]\partial_x[/tex].
 
ok so partial over the partial of x (x^2+y^2) = 2x, right?
so for what I'm doing the partial of x over the partial of z (x + iy) = (1)(partial of x over the partial of z) right?
 
FanofAFan said:
ok so partial over the partial of x (x^2+y^2) = 2x, right?
so for what I'm doing the partial of x over the partial of z (x + iy) = (1)(partial of x over the partial of z) right?

yes, that's right. The reason I'm asking you to write everything in terms of z and [tex]\bar{z}[/tex] is that it makes the calculations very very easy. (And it is very to useful to know the behavior of [tex]\partial_x[/tex], etc. in terms of z and [tex]\bar{z}[/tex] for future use)

What I want you to prove is

[tex]\Delta = \partial_x^2+\partial_y^2 = 4\partial_z\partial_{\bar{z}}[/tex]
[tex]\partial_x = \partial_z + \partial_{\bar{z}}[/tex]
[tex]\partial_y = i \partial_z - i \partial_{\bar{z}}[/tex]

Now [tex]W\Delta W = f(z)f(\bar{z})\Delta f(z)f(\bar{z}) = 4f(z)f(\bar{z})\partial_zf(z)\partial_{\bar{z}}f(\bar{z})[/tex]

Now [tex](\partial_x W)^2+(\partial_y W)^2 = ??[/tex] (show that it is equal to the above expression.)