Why is $ I = int_Omega frac{1}{z-zeta}Delta u(zeta) dzeta = -2i int_{partial Omega} frac{1}{z-zeta}partial...
I am unsure of some notation and also how a particular identity was derived. I read in a paper that because $Delta = 4 partial bar partial$ we have
$$
I = int_Omega frac{1}{z-zeta}Delta u(zeta) dzeta = -2i int_{partial Omega} frac{1}{z-zeta}partial u(zeta) dzeta.
$$
First of all, what is $partial bar partial$ and how is the Laplacian $Delta = 4 partial bar partial$?
Secondly, how has the identity been derived..it seems to be due to some integration by parts type formula, but what exactly?
integration complex-analysis holomorphic-functions greens-theorem
asked Nov 15 at 8:04
sonicboom
3,66282652
add a comment |
I am unsure of some notation and also how a particular identity was derived. I read in a paper that because $Delta = 4 partial bar partial$ we have
$$
I = int_Omega frac{1}{z-zeta}Delta u(zeta) dzeta = -2i int_{partial Omega} frac{1}{z-zeta}partial u(zeta) dzeta.
$$
First of all, what is $partial bar partial$ and how is the Laplacian $Delta = 4 partial bar partial$?
Secondly, how has the identity been derived..it seems to be due to some integration by parts type formula, but what exactly?
integration complex-analysis holomorphic-functions greens-theorem
asked Nov 15 at 8:04
sonicboom
3,66282652
It's a simple application of Gauß Theorem.
– Von Neumann
Nov 15 at 10:06
add a comment |
I am unsure of some notation and also how a particular identity was derived. I read in a paper that because $Delta = 4 partial bar partial$ we have
$$
I = int_Omega frac{1}{z-zeta}Delta u(zeta) dzeta = -2i int_{partial Omega} frac{1}{z-zeta}partial u(zeta) dzeta.
$$
First of all, what is $partial bar partial$ and how is the Laplacian $Delta = 4 partial bar partial$?
Secondly, how has the identity been derived..it seems to be due to some integration by parts type formula, but what exactly?
integration complex-analysis holomorphic-functions greens-theorem
asked Nov 15 at 8:04
sonicboom
3,66282652
I am unsure of some notation and also how a particular identity was derived. I read in a paper that because $Delta = 4 partial bar partial$ we have
$$
I = int_Omega frac{1}{z-zeta}Delta u(zeta) dzeta = -2i int_{partial Omega} frac{1}{z-zeta}partial u(zeta) dzeta.
$$
First of all, what is $partial bar partial$ and how is the Laplacian $Delta = 4 partial bar partial$?
Secondly, how has the identity been derived..it seems to be due to some integration by parts type formula, but what exactly?
integration complex-analysis holomorphic-functions greens-theorem
integration complex-analysis holomorphic-functions greens-theorem
asked Nov 15 at 8:04
sonicboom
3,66282652
asked Nov 15 at 8:04
sonicboom
3,66282652
asked Nov 15 at 8:04
sonicboom
3,66282652
asked Nov 15 at 8:04
sonicboom
3,66282652
asked Nov 15 at 8:04
sonicboom
3,66282652
3,66282652
It's a simple application of Gauß Theorem.
– Von Neumann
Nov 15 at 10:06
add a comment |
It's a simple application of Gauß Theorem.
– Von Neumann
Nov 15 at 10:06
It's a simple application of Gauß Theorem.
– Von Neumann
Nov 15 at 10:06
It's a simple application of Gauß Theorem.
– Von Neumann
Nov 15 at 10:06
add a comment |
1 Answer
1
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oldest
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Identifying the complex plane with $mathbb R^2$ with $z=x+iy$ the definition of $partial$ and $bar{partial}$ is
$$partial=frac{1}{2} left(partial_x+ipartial_y right)$$
$$bar{partial}=frac{1}{2} left(partial_x-ipartial_y right).$$
Then you indeed have
$$4partial bar{partial}= partial_x^2+partial_y^2 = Delta$$
Note that you can avoid using complex number as in $mathbb R^2$ a formulation can be:
$$partial=frac{1}{2} begin{pmatrix} partial_x \ partial_yend{pmatrix}=frac {1}{2}nabla$$
$$bar{partial}=frac{1}{2} begin{pmatrix} partial_x \ -partial_yend{pmatrix}=frac {1}{2} J nabla^perp=frac{1}{2} J operatorname{curl}$$
with $ J =begin{pmatrix} 0&-1\1&0end{pmatrix}$ correspond to $i$.
In these terms Green's theorem can be rewritten as
$$int_{partial Omega} g(zeta) d zeta = int_Omega 2i bar{partial} g(zeta) d zeta$$
We use this formula with $g(zeta)=frac{1}{z-zeta} partial u(zeta)$.
As $zeta mapsto frac{1}{z-zeta}$ is holomorphic you have $bar{partial} frac{1}{z-zeta}=0$ so
$$bar{partial} left( frac{1}{z-zeta} partial u(zeta) right)=frac{1}{z-zeta} bar{partial} partial u(zeta)=frac{1}{z-zeta}frac{1}{4} Delta u(zeta)$$
from where you obtain:
$$int_{partial Omega} frac{1}{z-zeta} partial u(zeta) d zeta= frac{2i}{4} int_Omega frac{1}{z-zeta} Delta u(zeta) dzeta$$
whic is the claimed result as $(i/2)^{-1}=-2i$.
answered Nov 15 at 10:31
Delta-u
5,3892618
Thanks..it seems there might be a sign error in the paper as it says that $4 hat partial partial = Delta$ which is the opposite to what you have.
– sonicboom
Nov 29 at 9:18
There is several small mistakes in my answer but it seems that the sign is the same. I will edit it to correct these mistakes to finish the computation.
– Delta-u
Nov 29 at 12:23
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Identifying the complex plane with $mathbb R^2$ with $z=x+iy$ the definition of $partial$ and $bar{partial}$ is
$$partial=frac{1}{2} left(partial_x+ipartial_y right)$$
$$bar{partial}=frac{1}{2} left(partial_x-ipartial_y right).$$
Then you indeed have
$$4partial bar{partial}= partial_x^2+partial_y^2 = Delta$$
Note that you can avoid using complex number as in $mathbb R^2$ a formulation can be:
$$partial=frac{1}{2} begin{pmatrix} partial_x \ partial_yend{pmatrix}=frac {1}{2}nabla$$
$$bar{partial}=frac{1}{2} begin{pmatrix} partial_x \ -partial_yend{pmatrix}=frac {1}{2} J nabla^perp=frac{1}{2} J operatorname{curl}$$
with $ J =begin{pmatrix} 0&-1\1&0end{pmatrix}$ correspond to $i$.
In these terms Green's theorem can be rewritten as
$$int_{partial Omega} g(zeta) d zeta = int_Omega 2i bar{partial} g(zeta) d zeta$$
We use this formula with $g(zeta)=frac{1}{z-zeta} partial u(zeta)$.
As $zeta mapsto frac{1}{z-zeta}$ is holomorphic you have $bar{partial} frac{1}{z-zeta}=0$ so
$$bar{partial} left( frac{1}{z-zeta} partial u(zeta) right)=frac{1}{z-zeta} bar{partial} partial u(zeta)=frac{1}{z-zeta}frac{1}{4} Delta u(zeta)$$
from where you obtain:
$$int_{partial Omega} frac{1}{z-zeta} partial u(zeta) d zeta= frac{2i}{4} int_Omega frac{1}{z-zeta} Delta u(zeta) dzeta$$
whic is the claimed result as $(i/2)^{-1}=-2i$.
answered Nov 15 at 10:31
Delta-u
5,3892618
Thanks..it seems there might be a sign error in the paper as it says that $4 hat partial partial = Delta$ which is the opposite to what you have.
– sonicboom
Nov 29 at 9:18
There is several small mistakes in my answer but it seems that the sign is the same. I will edit it to correct these mistakes to finish the computation.
– Delta-u
Nov 29 at 12:23
add a comment |
Identifying the complex plane with $mathbb R^2$ with $z=x+iy$ the definition of $partial$ and $bar{partial}$ is
$$partial=frac{1}{2} left(partial_x+ipartial_y right)$$
$$bar{partial}=frac{1}{2} left(partial_x-ipartial_y right).$$
Then you indeed have
$$4partial bar{partial}= partial_x^2+partial_y^2 = Delta$$
Note that you can avoid using complex number as in $mathbb R^2$ a formulation can be:
$$partial=frac{1}{2} begin{pmatrix} partial_x \ partial_yend{pmatrix}=frac {1}{2}nabla$$
$$bar{partial}=frac{1}{2} begin{pmatrix} partial_x \ -partial_yend{pmatrix}=frac {1}{2} J nabla^perp=frac{1}{2} J operatorname{curl}$$
with $ J =begin{pmatrix} 0&-1\1&0end{pmatrix}$ correspond to $i$.
In these terms Green's theorem can be rewritten as
$$int_{partial Omega} g(zeta) d zeta = int_Omega 2i bar{partial} g(zeta) d zeta$$
We use this formula with $g(zeta)=frac{1}{z-zeta} partial u(zeta)$.
As $zeta mapsto frac{1}{z-zeta}$ is holomorphic you have $bar{partial} frac{1}{z-zeta}=0$ so
$$bar{partial} left( frac{1}{z-zeta} partial u(zeta) right)=frac{1}{z-zeta} bar{partial} partial u(zeta)=frac{1}{z-zeta}frac{1}{4} Delta u(zeta)$$
from where you obtain:
$$int_{partial Omega} frac{1}{z-zeta} partial u(zeta) d zeta= frac{2i}{4} int_Omega frac{1}{z-zeta} Delta u(zeta) dzeta$$
whic is the claimed result as $(i/2)^{-1}=-2i$.
answered Nov 15 at 10:31
Delta-u
5,3892618
Thanks..it seems there might be a sign error in the paper as it says that $4 hat partial partial = Delta$ which is the opposite to what you have.
– sonicboom
Nov 29 at 9:18
There is several small mistakes in my answer but it seems that the sign is the same. I will edit it to correct these mistakes to finish the computation.
– Delta-u
Nov 29 at 12:23
add a comment |
Identifying the complex plane with $mathbb R^2$ with $z=x+iy$ the definition of $partial$ and $bar{partial}$ is
$$partial=frac{1}{2} left(partial_x+ipartial_y right)$$
$$bar{partial}=frac{1}{2} left(partial_x-ipartial_y right).$$
Then you indeed have
$$4partial bar{partial}= partial_x^2+partial_y^2 = Delta$$
Note that you can avoid using complex number as in $mathbb R^2$ a formulation can be:
$$partial=frac{1}{2} begin{pmatrix} partial_x \ partial_yend{pmatrix}=frac {1}{2}nabla$$
$$bar{partial}=frac{1}{2} begin{pmatrix} partial_x \ -partial_yend{pmatrix}=frac {1}{2} J nabla^perp=frac{1}{2} J operatorname{curl}$$
with $ J =begin{pmatrix} 0&-1\1&0end{pmatrix}$ correspond to $i$.
In these terms Green's theorem can be rewritten as
$$int_{partial Omega} g(zeta) d zeta = int_Omega 2i bar{partial} g(zeta) d zeta$$
We use this formula with $g(zeta)=frac{1}{z-zeta} partial u(zeta)$.
As $zeta mapsto frac{1}{z-zeta}$ is holomorphic you have $bar{partial} frac{1}{z-zeta}=0$ so
$$bar{partial} left( frac{1}{z-zeta} partial u(zeta) right)=frac{1}{z-zeta} bar{partial} partial u(zeta)=frac{1}{z-zeta}frac{1}{4} Delta u(zeta)$$
from where you obtain:
$$int_{partial Omega} frac{1}{z-zeta} partial u(zeta) d zeta= frac{2i}{4} int_Omega frac{1}{z-zeta} Delta u(zeta) dzeta$$
whic is the claimed result as $(i/2)^{-1}=-2i$.
answered Nov 15 at 10:31
Delta-u
5,3892618
Identifying the complex plane with $mathbb R^2$ with $z=x+iy$ the definition of $partial$ and $bar{partial}$ is
$$partial=frac{1}{2} left(partial_x+ipartial_y right)$$
$$bar{partial}=frac{1}{2} left(partial_x-ipartial_y right).$$
Then you indeed have
$$4partial bar{partial}= partial_x^2+partial_y^2 = Delta$$
Note that you can avoid using complex number as in $mathbb R^2$ a formulation can be:
$$partial=frac{1}{2} begin{pmatrix} partial_x \ partial_yend{pmatrix}=frac {1}{2}nabla$$
$$bar{partial}=frac{1}{2} begin{pmatrix} partial_x \ -partial_yend{pmatrix}=frac {1}{2} J nabla^perp=frac{1}{2} J operatorname{curl}$$
with $ J =begin{pmatrix} 0&-1\1&0end{pmatrix}$ correspond to $i$.
In these terms Green's theorem can be rewritten as
$$int_{partial Omega} g(zeta) d zeta = int_Omega 2i bar{partial} g(zeta) d zeta$$
We use this formula with $g(zeta)=frac{1}{z-zeta} partial u(zeta)$.
As $zeta mapsto frac{1}{z-zeta}$ is holomorphic you have $bar{partial} frac{1}{z-zeta}=0$ so
$$bar{partial} left( frac{1}{z-zeta} partial u(zeta) right)=frac{1}{z-zeta} bar{partial} partial u(zeta)=frac{1}{z-zeta}frac{1}{4} Delta u(zeta)$$
from where you obtain:
$$int_{partial Omega} frac{1}{z-zeta} partial u(zeta) d zeta= frac{2i}{4} int_Omega frac{1}{z-zeta} Delta u(zeta) dzeta$$
whic is the claimed result as $(i/2)^{-1}=-2i$.
answered Nov 15 at 10:31
Delta-u
5,3892618
edited Nov 29 at 12:28
answered Nov 15 at 10:31
Delta-u
5,3892618
answered Nov 15 at 10:31
Delta-u
5,3892618
answered Nov 15 at 10:31
Delta-u
5,3892618
5,3892618
Thanks..it seems there might be a sign error in the paper as it says that $4 hat partial partial = Delta$ which is the opposite to what you have.
– sonicboom
Nov 29 at 9:18
There is several small mistakes in my answer but it seems that the sign is the same. I will edit it to correct these mistakes to finish the computation.
– Delta-u
Nov 29 at 12:23
add a comment |
Thanks..it seems there might be a sign error in the paper as it says that $4 hat partial partial = Delta$ which is the opposite to what you have.
– sonicboom
Nov 29 at 9:18
There is several small mistakes in my answer but it seems that the sign is the same. I will edit it to correct these mistakes to finish the computation.
– Delta-u
Nov 29 at 12:23
Thanks..it seems there might be a sign error in the paper as it says that $4 hat partial partial = Delta$ which is the opposite to what you have.
– sonicboom
Nov 29 at 9:18
Thanks..it seems there might be a sign error in the paper as it says that $4 hat partial partial = Delta$ which is the opposite to what you have.
– sonicboom
Nov 29 at 9:18
There is several small mistakes in my answer but it seems that the sign is the same. I will edit it to correct these mistakes to finish the computation.
– Delta-u
Nov 29 at 12:23
There is several small mistakes in my answer but it seems that the sign is the same. I will edit it to correct these mistakes to finish the computation.
– Delta-u
Nov 29 at 12:23
add a comment |
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It's a simple application of Gauß Theorem.
– Von Neumann
Nov 15 at 10:06