Why is $ I = int_Omega frac{1}{z-zeta}Delta u(zeta) dzeta = -2i int_{partial Omega} frac{1}{z-zeta}partial...












1














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







share|cite|improve this question






















  • It's a simple application of Gauß Theorem.
    – Von Neumann
    Nov 15 at 10:06
















1














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







share|cite|improve this question






















  • It's a simple application of Gauß Theorem.
    – Von Neumann
    Nov 15 at 10:06














1












1








1







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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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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


















  • 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










1 Answer
1






active

oldest

votes


















1














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 $(i/2)^{-1}=-2i$.






share|cite|improve this answer























  • 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











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














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 $(i/2)^{-1}=-2i$.






share|cite|improve this answer























  • 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
















1














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 $(i/2)^{-1}=-2i$.






share|cite|improve this answer























  • 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














1












1








1






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 $(i/2)^{-1}=-2i$.






share|cite|improve this answer














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 $(i/2)^{-1}=-2i$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 29 at 12:28

























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


















  • 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


















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