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10:05 PM
$\nabla^2 G = \delta(x-\chi) \delta(y-\eta)$, $G(x,0;\chi,\eta)=0$...
what is the symmetry of this problem?
 
Yes, of course, $\log r$ is the only radially symmetric harmonic function on $\Bbb R^2-\{0\}$.
Turn the upper half plane into the unit disk and work there if you want.
 
oh I always forget to mention the upper half plane
$\nabla^2 G_{x,y}(x,y;\chi,\eta) = \delta(x-\chi) \delta(y-\eta)$, $G(x,0;\chi,\eta)=0$, $0 \le y, \eta$
@TedShifrin ^
oh wait you already knew of the upper half plane
that's why you explicitly mentioned "upper half plane"
wait can I use harmonic conjugate and instead find a meromorphic function that is purely imaginary on the real line and has a single singularity at $(\chi,\eta)$
 
10:22 PM
This is not stuff I think about ... And so I'm particularly rusty.
 
well it suffices to find one for $\chi+i\eta = i$ really
and suddenly we have a reflectional symmetry
wait did I say harmonic conjugate
it doesn't exist because my domain isn't simply connected, oh woe is me
 
Wait. I thought your domain was the upper half plane.
 
yes it is the upper half plane
 
So what's not simply connected?
 
but the pole makes things go awry
just like how $\ln r$ doesn't have a harmonic conjugate
 
10:31 PM
True.
 
hey that thing (upper plane minus i) is biholomorphic to the open disc right
no, the punctured open disc
 
No.
Yes.
 
so I only need to determine countably many numbers, yay
($\sum_{n \in \Bbb Z} a_n z^n$)
wait this is like Fourier series except they're not orthogonal
good grief, what is the explicit mapping
wait a Mobius transformation should do it
$T$ from punctured unit disc to upper half plane
-1 and 1 fixed, 0 to i
b/d=i, (a-b)/(c-d)=-1, (a+b)/(c+d)=1
b=i, d=1, a=c+1-i, c+1-2i=1-c, c=i, a=1
$T(z) = \dfrac{z+i}{iz+1}$
the inverse is $\dfrac{z-i}{-iz+1}$
so the function on the upper half plane is $\sum_{n\in\Bbb Z} a_n \left(\dfrac{z-i}{-iz+1}\right)^n$
what a lovely function
 
11:01 PM
the unit circle corresponds to the real axis
i.e. $\sum a_n z^n$ is imaginary for all $|z|=1$
but that means it's imaginary everywhere inside...
which means the whole function is zero to begin with??
time to sleep
 
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