How to find the harmonic conjugate of $v(x,y)=log((x-1)^2 +(y-2)^2)$?
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So I have found out $$frac{partial v}{partial x} = frac{2x-2}{(x-1)^2 +(y-2)^2}, frac{partial v}{partial y} = frac{2y-4}{(x-1)^2 +(y-2)^2}.$$
Using the Cauchy Riemann equations, I find:
$$frac{partial u}{partial x} = frac{2y-4}{(x-1)^2 +(y-2)^2}, frac{partial u}{partial y} = frac{2-2x}{(x-1)^2 +(y-2)^2}.$$
Then I integrate either one of $partial u/partial x$ or $partial u/partial y$ to then differentiate with respect to the other variable I just integrated with. However, when I integrate $partial u/partial x$ or $partial u/partial y$, I get two different functions: $2arctanfrac{(x-1)}{(y-2)} + g(y)$ and $-2arctanfrac{(y-2)}{(x-1)} + g(x)$, respectively.
What am I doing wrong? Can you have multiple harmonic conjuguates?
complex-analysis harmonic-functions
edited Nov 22 at 12:36
Ennar
14.3k32343
asked Nov 22 at 11:32
M. Calculator
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add a comment |
up vote
0
down vote
favorite
So I have found out $$frac{partial v}{partial x} = frac{2x-2}{(x-1)^2 +(y-2)^2}, frac{partial v}{partial y} = frac{2y-4}{(x-1)^2 +(y-2)^2}.$$
Using the Cauchy Riemann equations, I find:
$$frac{partial u}{partial x} = frac{2y-4}{(x-1)^2 +(y-2)^2}, frac{partial u}{partial y} = frac{2-2x}{(x-1)^2 +(y-2)^2}.$$
Then I integrate either one of $partial u/partial x$ or $partial u/partial y$ to then differentiate with respect to the other variable I just integrated with. However, when I integrate $partial u/partial x$ or $partial u/partial y$, I get two different functions: $2arctanfrac{(x-1)}{(y-2)} + g(y)$ and $-2arctanfrac{(y-2)}{(x-1)} + g(x)$, respectively.
What am I doing wrong? Can you have multiple harmonic conjuguates?
complex-analysis harmonic-functions
edited Nov 22 at 12:36
Ennar
14.3k32343
asked Nov 22 at 11:32
M. Calculator
376
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
So I have found out $$frac{partial v}{partial x} = frac{2x-2}{(x-1)^2 +(y-2)^2}, frac{partial v}{partial y} = frac{2y-4}{(x-1)^2 +(y-2)^2}.$$
Using the Cauchy Riemann equations, I find:
$$frac{partial u}{partial x} = frac{2y-4}{(x-1)^2 +(y-2)^2}, frac{partial u}{partial y} = frac{2-2x}{(x-1)^2 +(y-2)^2}.$$
Then I integrate either one of $partial u/partial x$ or $partial u/partial y$ to then differentiate with respect to the other variable I just integrated with. However, when I integrate $partial u/partial x$ or $partial u/partial y$, I get two different functions: $2arctanfrac{(x-1)}{(y-2)} + g(y)$ and $-2arctanfrac{(y-2)}{(x-1)} + g(x)$, respectively.
What am I doing wrong? Can you have multiple harmonic conjuguates?
complex-analysis harmonic-functions
edited Nov 22 at 12:36
Ennar
14.3k32343
asked Nov 22 at 11:32
M. Calculator
376
So I have found out $$frac{partial v}{partial x} = frac{2x-2}{(x-1)^2 +(y-2)^2}, frac{partial v}{partial y} = frac{2y-4}{(x-1)^2 +(y-2)^2}.$$
Using the Cauchy Riemann equations, I find:
$$frac{partial u}{partial x} = frac{2y-4}{(x-1)^2 +(y-2)^2}, frac{partial u}{partial y} = frac{2-2x}{(x-1)^2 +(y-2)^2}.$$
Then I integrate either one of $partial u/partial x$ or $partial u/partial y$ to then differentiate with respect to the other variable I just integrated with. However, when I integrate $partial u/partial x$ or $partial u/partial y$, I get two different functions: $2arctanfrac{(x-1)}{(y-2)} + g(y)$ and $-2arctanfrac{(y-2)}{(x-1)} + g(x)$, respectively.
What am I doing wrong? Can you have multiple harmonic conjuguates?
complex-analysis harmonic-functions
complex-analysis harmonic-functions
edited Nov 22 at 12:36
Ennar
14.3k32343
asked Nov 22 at 11:32
M. Calculator
376
edited Nov 22 at 12:36
Ennar
14.3k32343
asked Nov 22 at 11:32
M. Calculator
376
edited Nov 22 at 12:36
Ennar
14.3k32343
edited Nov 22 at 12:36
Ennar
14.3k32343
edited Nov 22 at 12:36
Ennar
14.3k32343
14.3k32343
asked Nov 22 at 11:32
M. Calculator
376
asked Nov 22 at 11:32
M. Calculator
376
asked Nov 22 at 11:32
M. Calculator
376
376
add a comment |
add a comment |
1 Answer
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You will not be able to find an harmonic conjugate of $v$ since there is none. That is, there is no function $ucolonmathbb{C}setminus{1+2i}longrightarrowmathbb C$ which is an harmonic conjugate of $v$. You will find a proof here.
answered Nov 22 at 11:41
José Carlos Santos
146k22117217
1
I have deleted my first sentence. On the other hand, I thought that your problem was to find an harmonic conjugate $u$ of $v$. Am I wrong?
– José Carlos Santos
Nov 22 at 11:51
1
No. The fact that $v$ is harmonic only assures that a harmonic conjugate exists localy. But your function is a standard example of function without a global harmonic conjugate.
– José Carlos Santos
Nov 22 at 11:56
1
For each $(x,y)in D_v$ there is an open ball $B$ centered at that point such that $v|_B$ has a harmonic conjugate; that's the meaning of the fact that locally $v$ has a harmonic conjugate. However, there is no harmonic conjugate of $v$ whose domain is $D_v$; that's what it means that there is no harmonic conjugate globally.
– José Carlos Santos
Nov 22 at 12:02
1
Actually, $1+2i$ is a bad choice, since $1+2i$ is the only point outside the domain of $v$. There is no global harmonic conjugate of $v$, but there is one near every point of its domain.
– José Carlos Santos
Nov 22 at 12:26
1
I have no more ways of telling you that $v$ has no harmonic conjugate. I'm off.
– José Carlos Santos
Nov 22 at 12:37
|
show 6 more comments
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
up vote
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down vote
You will not be able to find an harmonic conjugate of $v$ since there is none. That is, there is no function $ucolonmathbb{C}setminus{1+2i}longrightarrowmathbb C$ which is an harmonic conjugate of $v$. You will find a proof here.
answered Nov 22 at 11:41
José Carlos Santos
146k22117217
1
I have deleted my first sentence. On the other hand, I thought that your problem was to find an harmonic conjugate $u$ of $v$. Am I wrong?
– José Carlos Santos
Nov 22 at 11:51
1
No. The fact that $v$ is harmonic only assures that a harmonic conjugate exists localy. But your function is a standard example of function without a global harmonic conjugate.
– José Carlos Santos
Nov 22 at 11:56
1
For each $(x,y)in D_v$ there is an open ball $B$ centered at that point such that $v|_B$ has a harmonic conjugate; that's the meaning of the fact that locally $v$ has a harmonic conjugate. However, there is no harmonic conjugate of $v$ whose domain is $D_v$; that's what it means that there is no harmonic conjugate globally.
– José Carlos Santos
Nov 22 at 12:02
1
Actually, $1+2i$ is a bad choice, since $1+2i$ is the only point outside the domain of $v$. There is no global harmonic conjugate of $v$, but there is one near every point of its domain.
– José Carlos Santos
Nov 22 at 12:26
1
I have no more ways of telling you that $v$ has no harmonic conjugate. I'm off.
– José Carlos Santos
Nov 22 at 12:37
|
show 6 more comments
up vote
1
down vote
You will not be able to find an harmonic conjugate of $v$ since there is none. That is, there is no function $ucolonmathbb{C}setminus{1+2i}longrightarrowmathbb C$ which is an harmonic conjugate of $v$. You will find a proof here.
answered Nov 22 at 11:41
José Carlos Santos
146k22117217
1
I have deleted my first sentence. On the other hand, I thought that your problem was to find an harmonic conjugate $u$ of $v$. Am I wrong?
– José Carlos Santos
Nov 22 at 11:51
1
No. The fact that $v$ is harmonic only assures that a harmonic conjugate exists localy. But your function is a standard example of function without a global harmonic conjugate.
– José Carlos Santos
Nov 22 at 11:56
1
For each $(x,y)in D_v$ there is an open ball $B$ centered at that point such that $v|_B$ has a harmonic conjugate; that's the meaning of the fact that locally $v$ has a harmonic conjugate. However, there is no harmonic conjugate of $v$ whose domain is $D_v$; that's what it means that there is no harmonic conjugate globally.
– José Carlos Santos
Nov 22 at 12:02
1
Actually, $1+2i$ is a bad choice, since $1+2i$ is the only point outside the domain of $v$. There is no global harmonic conjugate of $v$, but there is one near every point of its domain.
– José Carlos Santos
Nov 22 at 12:26
1
I have no more ways of telling you that $v$ has no harmonic conjugate. I'm off.
– José Carlos Santos
Nov 22 at 12:37
|
show 6 more comments
up vote
1
down vote
up vote
1
down vote
You will not be able to find an harmonic conjugate of $v$ since there is none. That is, there is no function $ucolonmathbb{C}setminus{1+2i}longrightarrowmathbb C$ which is an harmonic conjugate of $v$. You will find a proof here.
answered Nov 22 at 11:41
José Carlos Santos
146k22117217
You will not be able to find an harmonic conjugate of $v$ since there is none. That is, there is no function $ucolonmathbb{C}setminus{1+2i}longrightarrowmathbb C$ which is an harmonic conjugate of $v$. You will find a proof here.
answered Nov 22 at 11:41
José Carlos Santos
146k22117217
edited Nov 22 at 11:49
answered Nov 22 at 11:41
José Carlos Santos
146k22117217
answered Nov 22 at 11:41
José Carlos Santos
146k22117217
answered Nov 22 at 11:41
José Carlos Santos
146k22117217
146k22117217
1
I have deleted my first sentence. On the other hand, I thought that your problem was to find an harmonic conjugate $u$ of $v$. Am I wrong?
– José Carlos Santos
Nov 22 at 11:51
1
No. The fact that $v$ is harmonic only assures that a harmonic conjugate exists localy. But your function is a standard example of function without a global harmonic conjugate.
– José Carlos Santos
Nov 22 at 11:56
1
For each $(x,y)in D_v$ there is an open ball $B$ centered at that point such that $v|_B$ has a harmonic conjugate; that's the meaning of the fact that locally $v$ has a harmonic conjugate. However, there is no harmonic conjugate of $v$ whose domain is $D_v$; that's what it means that there is no harmonic conjugate globally.
– José Carlos Santos
Nov 22 at 12:02
1
Actually, $1+2i$ is a bad choice, since $1+2i$ is the only point outside the domain of $v$. There is no global harmonic conjugate of $v$, but there is one near every point of its domain.
– José Carlos Santos
Nov 22 at 12:26
1
I have no more ways of telling you that $v$ has no harmonic conjugate. I'm off.
– José Carlos Santos
Nov 22 at 12:37
|
show 6 more comments
1
I have deleted my first sentence. On the other hand, I thought that your problem was to find an harmonic conjugate $u$ of $v$. Am I wrong?
– José Carlos Santos
Nov 22 at 11:51
1
No. The fact that $v$ is harmonic only assures that a harmonic conjugate exists localy. But your function is a standard example of function without a global harmonic conjugate.
– José Carlos Santos
Nov 22 at 11:56
1
For each $(x,y)in D_v$ there is an open ball $B$ centered at that point such that $v|_B$ has a harmonic conjugate; that's the meaning of the fact that locally $v$ has a harmonic conjugate. However, there is no harmonic conjugate of $v$ whose domain is $D_v$; that's what it means that there is no harmonic conjugate globally.
– José Carlos Santos
Nov 22 at 12:02
1
Actually, $1+2i$ is a bad choice, since $1+2i$ is the only point outside the domain of $v$. There is no global harmonic conjugate of $v$, but there is one near every point of its domain.
– José Carlos Santos
Nov 22 at 12:26
1
I have no more ways of telling you that $v$ has no harmonic conjugate. I'm off.
– José Carlos Santos
Nov 22 at 12:37
1
1
I have deleted my first sentence. On the other hand, I thought that your problem was to find an harmonic conjugate $u$ of $v$. Am I wrong?
– José Carlos Santos
Nov 22 at 11:51
I have deleted my first sentence. On the other hand, I thought that your problem was to find an harmonic conjugate $u$ of $v$. Am I wrong?
– José Carlos Santos
Nov 22 at 11:51
1
1
No. The fact that $v$ is harmonic only assures that a harmonic conjugate exists localy. But your function is a standard example of function without a global harmonic conjugate.
– José Carlos Santos
Nov 22 at 11:56
No. The fact that $v$ is harmonic only assures that a harmonic conjugate exists localy. But your function is a standard example of function without a global harmonic conjugate.
– José Carlos Santos
Nov 22 at 11:56
1
1
For each $(x,y)in D_v$ there is an open ball $B$ centered at that point such that $v|_B$ has a harmonic conjugate; that's the meaning of the fact that locally $v$ has a harmonic conjugate. However, there is no harmonic conjugate of $v$ whose domain is $D_v$; that's what it means that there is no harmonic conjugate globally.
– José Carlos Santos
Nov 22 at 12:02
For each $(x,y)in D_v$ there is an open ball $B$ centered at that point such that $v|_B$ has a harmonic conjugate; that's the meaning of the fact that locally $v$ has a harmonic conjugate. However, there is no harmonic conjugate of $v$ whose domain is $D_v$; that's what it means that there is no harmonic conjugate globally.
– José Carlos Santos
Nov 22 at 12:02
1
1
Actually, $1+2i$ is a bad choice, since $1+2i$ is the only point outside the domain of $v$. There is no global harmonic conjugate of $v$, but there is one near every point of its domain.
– José Carlos Santos
Nov 22 at 12:26
Actually, $1+2i$ is a bad choice, since $1+2i$ is the only point outside the domain of $v$. There is no global harmonic conjugate of $v$, but there is one near every point of its domain.
– José Carlos Santos
Nov 22 at 12:26
1
1
I have no more ways of telling you that $v$ has no harmonic conjugate. I'm off.
– José Carlos Santos
Nov 22 at 12:37
I have no more ways of telling you that $v$ has no harmonic conjugate. I'm off.
– José Carlos Santos
Nov 22 at 12:37
|
show 6 more comments
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