Harmonic conjugate of $ln(|z|)$
Find the harmonic conjugate of $u$. $u = u(z) = ln(|z|)$ so $u(z) = ln(sqrt{x^2 + y^2})$
I am trying to find now its harmonic conjugate I did all the math:
I got two solutions though.
One is $v(z) = arctan(y/x) + C$ if I solve $partial u/partial x = -partial v/partial y$ & other is
$v(z) = - arctan(x/y) + C$ if I solved $partial v/partial y = partial u/partial x$
can I have two solutions, or is only one solution correct?
complex-analysis
edited Feb 23 '15 at 0:31
Stephen Montgomery-Smith
17.7k12247
asked Oct 9 '13 at 5:22
Illustionist
379211
add a comment |
Find the harmonic conjugate of $u$. $u = u(z) = ln(|z|)$ so $u(z) = ln(sqrt{x^2 + y^2})$
I am trying to find now its harmonic conjugate I did all the math:
I got two solutions though.
One is $v(z) = arctan(y/x) + C$ if I solve $partial u/partial x = -partial v/partial y$ & other is
$v(z) = - arctan(x/y) + C$ if I solved $partial v/partial y = partial u/partial x$
can I have two solutions, or is only one solution correct?
complex-analysis
edited Feb 23 '15 at 0:31
Stephen Montgomery-Smith
17.7k12247
asked Oct 9 '13 at 5:22
Illustionist
379211
Since I just answered another, I thought I would see if I could answer this one for you as well.
– Stephen Montgomery-Smith
Feb 23 '15 at 0:30
add a comment |
Find the harmonic conjugate of $u$. $u = u(z) = ln(|z|)$ so $u(z) = ln(sqrt{x^2 + y^2})$
I am trying to find now its harmonic conjugate I did all the math:
I got two solutions though.
One is $v(z) = arctan(y/x) + C$ if I solve $partial u/partial x = -partial v/partial y$ & other is
$v(z) = - arctan(x/y) + C$ if I solved $partial v/partial y = partial u/partial x$
can I have two solutions, or is only one solution correct?
complex-analysis
edited Feb 23 '15 at 0:31
Stephen Montgomery-Smith
17.7k12247
asked Oct 9 '13 at 5:22
Illustionist
379211
Find the harmonic conjugate of $u$. $u = u(z) = ln(|z|)$ so $u(z) = ln(sqrt{x^2 + y^2})$
I am trying to find now its harmonic conjugate I did all the math:
I got two solutions though.
One is $v(z) = arctan(y/x) + C$ if I solve $partial u/partial x = -partial v/partial y$ & other is
$v(z) = - arctan(x/y) + C$ if I solved $partial v/partial y = partial u/partial x$
can I have two solutions, or is only one solution correct?
complex-analysis
complex-analysis
edited Feb 23 '15 at 0:31
Stephen Montgomery-Smith
17.7k12247
asked Oct 9 '13 at 5:22
Illustionist
379211
edited Feb 23 '15 at 0:31
Stephen Montgomery-Smith
17.7k12247
asked Oct 9 '13 at 5:22
Illustionist
379211
edited Feb 23 '15 at 0:31
Stephen Montgomery-Smith
17.7k12247
edited Feb 23 '15 at 0:31
Stephen Montgomery-Smith
17.7k12247
edited Feb 23 '15 at 0:31
Stephen Montgomery-Smith
17.7k12247
17.7k12247
asked Oct 9 '13 at 5:22
Illustionist
379211
asked Oct 9 '13 at 5:22
Illustionist
379211
asked Oct 9 '13 at 5:22
Illustionist
379211
379211
Since I just answered another, I thought I would see if I could answer this one for you as well.
– Stephen Montgomery-Smith
Feb 23 '15 at 0:30
add a comment |
Since I just answered another, I thought I would see if I could answer this one for you as well.
– Stephen Montgomery-Smith
Feb 23 '15 at 0:30
Since I just answered another, I thought I would see if I could answer this one for you as well.
– Stephen Montgomery-Smith
Feb 23 '15 at 0:30
Since I just answered another, I thought I would see if I could answer this one for you as well.
– Stephen Montgomery-Smith
Feb 23 '15 at 0:30
add a comment |
2 Answers
2
active
oldest
votes
$arctan(y/x) + arctan(x/y) = pm fracpi2 $. So in essence, your two solutions are the same.
Another way to get your solution: $ln(z)$ is analytic, and if $z = r e^{itheta}$, then $ln(r e^{itheta}) = ln r + itheta$. And $theta$ is the angle $z$ makes to the $x$-axis, that is, $arctan(y/x)$. You can get the other solutions by considering, for example, $ln(iz)$, or more generally $ln(az)$ where $|a| = 1$.
answered Feb 23 '15 at 0:23
Stephen Montgomery-Smith
17.7k12247
add a comment |
A harmonic function is a twice differentiable function that satisfies Laplace's equation. Recall that for a function to be differentiable, it must first be continuous.
As you found out by going through the calculations, the harmonic conjugate $v(z)$, if it exists, must be either $v(z) = arctan(y/x) + C_1(x)$ (where $C_1$ is a constant with respect to $y$ but is possibly a non-constant function of $x$) or $v(z) = arctan(x/y) + C_2(y)$.
However, neither choice is continuous on $mathbb{R}^2$ or $mathbb{C}$, so neither choice is harmonic. Therefore $u(z)$ does not have a harmonic conjugate.
To see that neither choice is continuous, consider the limit $(x, y) to (0, 0)$ of $arctan(y/x)$ and $arctan(x/y)$ along the lines $y = x$ and $y = -x$. Along the first line, they are both equal to $arctan(1) = sqrt{3}$; along the second line they are both equal to $arctan(-1) = -sqrt{3}$.
answered Oct 9 '13 at 7:33
Kyle
1,092717
Well, $u(z)$ itself is discontinuous at $0$, so it is not surprising that $v$ is discontinuous there. You will get a harmonic conjugate on any simply connected domain where $u$ is defined.
– GEdgar
Feb 23 '15 at 0:37
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
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active
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active
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votes
$arctan(y/x) + arctan(x/y) = pm fracpi2 $. So in essence, your two solutions are the same.
Another way to get your solution: $ln(z)$ is analytic, and if $z = r e^{itheta}$, then $ln(r e^{itheta}) = ln r + itheta$. And $theta$ is the angle $z$ makes to the $x$-axis, that is, $arctan(y/x)$. You can get the other solutions by considering, for example, $ln(iz)$, or more generally $ln(az)$ where $|a| = 1$.
answered Feb 23 '15 at 0:23
Stephen Montgomery-Smith
17.7k12247
add a comment |
$arctan(y/x) + arctan(x/y) = pm fracpi2 $. So in essence, your two solutions are the same.
Another way to get your solution: $ln(z)$ is analytic, and if $z = r e^{itheta}$, then $ln(r e^{itheta}) = ln r + itheta$. And $theta$ is the angle $z$ makes to the $x$-axis, that is, $arctan(y/x)$. You can get the other solutions by considering, for example, $ln(iz)$, or more generally $ln(az)$ where $|a| = 1$.
answered Feb 23 '15 at 0:23
Stephen Montgomery-Smith
17.7k12247
add a comment |
$arctan(y/x) + arctan(x/y) = pm fracpi2 $. So in essence, your two solutions are the same.
Another way to get your solution: $ln(z)$ is analytic, and if $z = r e^{itheta}$, then $ln(r e^{itheta}) = ln r + itheta$. And $theta$ is the angle $z$ makes to the $x$-axis, that is, $arctan(y/x)$. You can get the other solutions by considering, for example, $ln(iz)$, or more generally $ln(az)$ where $|a| = 1$.
answered Feb 23 '15 at 0:23
Stephen Montgomery-Smith
17.7k12247
$arctan(y/x) + arctan(x/y) = pm fracpi2 $. So in essence, your two solutions are the same.
Another way to get your solution: $ln(z)$ is analytic, and if $z = r e^{itheta}$, then $ln(r e^{itheta}) = ln r + itheta$. And $theta$ is the angle $z$ makes to the $x$-axis, that is, $arctan(y/x)$. You can get the other solutions by considering, for example, $ln(iz)$, or more generally $ln(az)$ where $|a| = 1$.
answered Feb 23 '15 at 0:23
Stephen Montgomery-Smith
17.7k12247
edited Feb 23 '15 at 0:28
answered Feb 23 '15 at 0:23
Stephen Montgomery-Smith
17.7k12247
answered Feb 23 '15 at 0:23
Stephen Montgomery-Smith
17.7k12247
answered Feb 23 '15 at 0:23
Stephen Montgomery-Smith
17.7k12247
17.7k12247
add a comment |
add a comment |
A harmonic function is a twice differentiable function that satisfies Laplace's equation. Recall that for a function to be differentiable, it must first be continuous.
As you found out by going through the calculations, the harmonic conjugate $v(z)$, if it exists, must be either $v(z) = arctan(y/x) + C_1(x)$ (where $C_1$ is a constant with respect to $y$ but is possibly a non-constant function of $x$) or $v(z) = arctan(x/y) + C_2(y)$.
However, neither choice is continuous on $mathbb{R}^2$ or $mathbb{C}$, so neither choice is harmonic. Therefore $u(z)$ does not have a harmonic conjugate.
To see that neither choice is continuous, consider the limit $(x, y) to (0, 0)$ of $arctan(y/x)$ and $arctan(x/y)$ along the lines $y = x$ and $y = -x$. Along the first line, they are both equal to $arctan(1) = sqrt{3}$; along the second line they are both equal to $arctan(-1) = -sqrt{3}$.
answered Oct 9 '13 at 7:33
Kyle
1,092717
Well, $u(z)$ itself is discontinuous at $0$, so it is not surprising that $v$ is discontinuous there. You will get a harmonic conjugate on any simply connected domain where $u$ is defined.
– GEdgar
Feb 23 '15 at 0:37
add a comment |
A harmonic function is a twice differentiable function that satisfies Laplace's equation. Recall that for a function to be differentiable, it must first be continuous.
As you found out by going through the calculations, the harmonic conjugate $v(z)$, if it exists, must be either $v(z) = arctan(y/x) + C_1(x)$ (where $C_1$ is a constant with respect to $y$ but is possibly a non-constant function of $x$) or $v(z) = arctan(x/y) + C_2(y)$.
However, neither choice is continuous on $mathbb{R}^2$ or $mathbb{C}$, so neither choice is harmonic. Therefore $u(z)$ does not have a harmonic conjugate.
To see that neither choice is continuous, consider the limit $(x, y) to (0, 0)$ of $arctan(y/x)$ and $arctan(x/y)$ along the lines $y = x$ and $y = -x$. Along the first line, they are both equal to $arctan(1) = sqrt{3}$; along the second line they are both equal to $arctan(-1) = -sqrt{3}$.
answered Oct 9 '13 at 7:33
Kyle
1,092717
Well, $u(z)$ itself is discontinuous at $0$, so it is not surprising that $v$ is discontinuous there. You will get a harmonic conjugate on any simply connected domain where $u$ is defined.
– GEdgar
Feb 23 '15 at 0:37
add a comment |
A harmonic function is a twice differentiable function that satisfies Laplace's equation. Recall that for a function to be differentiable, it must first be continuous.
As you found out by going through the calculations, the harmonic conjugate $v(z)$, if it exists, must be either $v(z) = arctan(y/x) + C_1(x)$ (where $C_1$ is a constant with respect to $y$ but is possibly a non-constant function of $x$) or $v(z) = arctan(x/y) + C_2(y)$.
However, neither choice is continuous on $mathbb{R}^2$ or $mathbb{C}$, so neither choice is harmonic. Therefore $u(z)$ does not have a harmonic conjugate.
To see that neither choice is continuous, consider the limit $(x, y) to (0, 0)$ of $arctan(y/x)$ and $arctan(x/y)$ along the lines $y = x$ and $y = -x$. Along the first line, they are both equal to $arctan(1) = sqrt{3}$; along the second line they are both equal to $arctan(-1) = -sqrt{3}$.
answered Oct 9 '13 at 7:33
Kyle
1,092717
A harmonic function is a twice differentiable function that satisfies Laplace's equation. Recall that for a function to be differentiable, it must first be continuous.
As you found out by going through the calculations, the harmonic conjugate $v(z)$, if it exists, must be either $v(z) = arctan(y/x) + C_1(x)$ (where $C_1$ is a constant with respect to $y$ but is possibly a non-constant function of $x$) or $v(z) = arctan(x/y) + C_2(y)$.
However, neither choice is continuous on $mathbb{R}^2$ or $mathbb{C}$, so neither choice is harmonic. Therefore $u(z)$ does not have a harmonic conjugate.
To see that neither choice is continuous, consider the limit $(x, y) to (0, 0)$ of $arctan(y/x)$ and $arctan(x/y)$ along the lines $y = x$ and $y = -x$. Along the first line, they are both equal to $arctan(1) = sqrt{3}$; along the second line they are both equal to $arctan(-1) = -sqrt{3}$.
answered Oct 9 '13 at 7:33
Kyle
1,092717
answered Oct 9 '13 at 7:33
Kyle
1,092717
answered Oct 9 '13 at 7:33
Kyle
1,092717
answered Oct 9 '13 at 7:33
Kyle
1,092717
1,092717
Well, $u(z)$ itself is discontinuous at $0$, so it is not surprising that $v$ is discontinuous there. You will get a harmonic conjugate on any simply connected domain where $u$ is defined.
– GEdgar
Feb 23 '15 at 0:37
add a comment |
Well, $u(z)$ itself is discontinuous at $0$, so it is not surprising that $v$ is discontinuous there. You will get a harmonic conjugate on any simply connected domain where $u$ is defined.
– GEdgar
Feb 23 '15 at 0:37
Well, $u(z)$ itself is discontinuous at $0$, so it is not surprising that $v$ is discontinuous there. You will get a harmonic conjugate on any simply connected domain where $u$ is defined.
– GEdgar
Feb 23 '15 at 0:37
Well, $u(z)$ itself is discontinuous at $0$, so it is not surprising that $v$ is discontinuous there. You will get a harmonic conjugate on any simply connected domain where $u$ is defined.
– GEdgar
Feb 23 '15 at 0:37
add a comment |
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Since I just answered another, I thought I would see if I could answer this one for you as well.
– Stephen Montgomery-Smith
Feb 23 '15 at 0:30