Harmonic Functions (Complex Analysis)

I've been revising for 5 hours straight and for some reason a few little things have slipped me up:

Spoiler

How is it that both of these MVPs can be rewritten in the forms they are given? I have thought a direct change of variables but maybe I am being silly, I haven't been able to get them. They make sense intuitvely but I cannot seem to do it mathematically.. I am probably missing something very obvious.

EDIT: I just got the second MVP but still not the first one..

Another question:

Spoiler

what is the reason for taking the laplacian of the log of x?

The proof goes on and it makes sense from there on but I just have no idea why the laplacian of log of x was chosen.

Apologies if these seem like silly questions, there were things just skimmed over during lectures and I didn't think much of them at the time!

Thank you for any help :smile:

(edited 7 years ago)

Reply 1

Math12345

14

I am sort of guessing this:

Let

y = x + rn(\theta)

where

\theta \in [0,2 \pi]

dy/d \theta = r(- \sin \theta, \cos \theta)

dS_y = r \sqrt {(- \sin \theta)^2 + (\cos \theta)^2} d \theta

So

dS_y = r d \theta

(using

sin^2(x)+cos^2(x)=1

) and the r will cancel out.

For the laplacian question: see what conditions are needed (hence log can be used).

(edited 7 years ago)

Reply 2

Farhan.Hanif93

18

Original post by WannabeBeAST

EDIT: I just got the second MVP but still not the first one..

The first is straightforward by noting that

\displaystyle\int_C fds = \displaystyle\int_a^b f\left| \dfrac{d\mathbf{r}}{du}\right| du

Where

ds

is the scalar line element defined via

d\mathbf{r} = \hat{\mathbf{t}}ds

and

u

parameterises the curve

C

s.t.

a,b

are the suitable limits.

Another question:

Spoiler

what is the reason for taking the laplacian of the log of x?

The proof goes on and it makes sense from there on but I just have no idea why the laplacian of log of x was chosen.

Apologies if these seem like silly questions, there were things just skimmed over during lectures and I didn't think much of them at the time!

Thank you for any help :smile:

I think it may be unclear because the key identity is written in an unhelpful way. Moreso than the divergence theorem, the intuition behind the proof really relies on Green's second identity (which is a corollary): For suitable functions

f, g

, we have (in the notation of your notes)

\displaystyle\int_{\Omega_0} (f(x)\Delta{g(x)} - g(x)\Delta{f(x)})dx = \displaystyle\int_{\partial \Omega_0} (f(x)\nabla g(x) - g(x)\nabla f(x))\cdot \hat{n} dS(x)

Looking at it from here and keeping in mind that you want to reduce to the first form of the mean value property, it's sensible to choose one of these to be

u(x+y)

-

f

, say. This leaves us with the choice of

g

and collapses our above identity to:

\displaystyle\int_{\Omega_0} u(x+y)\Delta{g(y)} dy = \displaystyle\int_{\partial \Omega_0} u(x+y)\nabla g(y) \cdot \hat{n} dS(y)

- \displaystyle\int_{\partial \Omega_0} g(y)\nabla u(x+y) \cdot \hat{n} dS(y)

It's clear that if we force

g

be harmonic on

\Omega_0

and constant on

\partial \Omega_0

, the LHS and the second integral on the RHS both vanish, leaving us with:

\displaystyle\int_{\partial \Omega_0} u(x+y)\nabla g(y) \cdot \hat{n} dS(y) = 0

Defining

\Omega_0

as the proof suggests allows us to then see that, if

\nabla g(y) \cdot \hat{n}

is constant on the two boundaries of the annulus individually, the desired result will follow.

Hence we require a solution to Laplace's equation with this set of Neumann boundary conditions. It turns out that