Harmonic Functions With a Pole at the Origin Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Reflection principle for harmonic functionsBounded (from below) harmonic functions from $mathbb R^2 setminus 0$Show that if $g$ is nonconstant holomorphic and $f$ is harmonic such that $fg$ is harmonic, then $f$ is holomorphic.Limit of bounded harmonic functions is harmonicHarmonic functions and mean value propertyProve that a harmonic function is an open map.Understanding “a harmonicity argument”If $U$ is simply connected and $u: UtoBbb R$ is harmonic then it have a conjugate in $U$Showing $f=u+iv$ satisfies $f'(x)=u_x(x,0)-iu_y(x,0)$ for real $x$.Linear harmonic functions
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Harmonic Functions With a Pole at the Origin
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Reflection principle for harmonic functionsBounded (from below) harmonic functions from $mathbb R^2 setminus 0$Show that if $g$ is nonconstant holomorphic and $f$ is harmonic such that $fg$ is harmonic, then $f$ is holomorphic.Limit of bounded harmonic functions is harmonicHarmonic functions and mean value propertyProve that a harmonic function is an open map.Understanding “a harmonicity argument”If $U$ is simply connected and $u: UtoBbb R$ is harmonic then it have a conjugate in $U$Showing $f=u+iv$ satisfies $f'(x)=u_x(x,0)-iu_y(x,0)$ for real $x$.Linear harmonic functions
$begingroup$
I'm trying to solve the following problem:
Suppose that $u:mathbb D setminus0 to mathbb R$ is harmonic and that $lim_zto 0 u(z)=infty$.
Show that $u$ can be written as $$u(z)=beta ln |z| + v(z) $$ where $betain mathbb R setminus 0$ and $v:mathbb D to mathbb R $ is harmonic in $mathbb D$ (Here $mathbb D = z $).
I was also given the following hint:
Hint: Show that the residue of $(u_x - i u_y) $ is real.
I believe that the solution sould be related to Dirichlet's problem, and especially to Poisson's formula, but I don't understand how to use those to solve it.
Any help would will be appreciated!
complex-analysis harmonic-functions
asked Apr 1 at 18:15
Yarin LuhmanyYarin Luhmany
512
$endgroup$
add a comment |
$begingroup$
I'm trying to solve the following problem:
Suppose that $u:mathbb D setminus0 to mathbb R$ is harmonic and that $lim_zto 0 u(z)=infty$.
Show that $u$ can be written as $$u(z)=beta ln |z| + v(z) $$ where $betain mathbb R setminus 0$ and $v:mathbb D to mathbb R $ is harmonic in $mathbb D$ (Here $mathbb D = z $).
I was also given the following hint:
Hint: Show that the residue of $(u_x - i u_y) $ is real.
I believe that the solution sould be related to Dirichlet's problem, and especially to Poisson's formula, but I don't understand how to use those to solve it.
Any help would will be appreciated!
complex-analysis harmonic-functions
asked Apr 1 at 18:15
Yarin LuhmanyYarin Luhmany
512
$endgroup$
1
$begingroup$
On any simply connected open set $Usubset 0<$, $u(z) = Re(f(z))$ where $f$ is analytic on $U$, and the analytic continuation of $f$ is analytic over any curve $subset 0<$, and over closed curves, the continuation $f_gamma$ differs off $f$ by a purely imaginary constant. Let $A = lim_t to 0 f(e^2i pi (1-t))- f(e^2i pi t)$ which is purely imaginary. Then $g(z)=f(z)- A fraclog z2ipi$ is analytic on $0 < |z| < 1$.
$endgroup$
– reuns
Apr 1 at 23:18
1
$begingroup$
From there look at the 3 cases : $g$ has an essential singularity, a pole, a removable singularity, and show only in the latter case $lim_z to 0 |u(z)| = infty$.
$endgroup$
– reuns
Apr 1 at 23:18
$begingroup$
Thank you for your answer! Why can we use analytic continuation (for $f$)? Also I'd love an explantion for the definition of $A$ (we use $f(e^2pi i t)$, but $f$ is only defined on $Usubseteq 0<$....)
$endgroup$
– Yarin Luhmany
Apr 2 at 5:33
1
$begingroup$
From the Poisson kernel you can construct $f$ analytic on $0<|z|<1, z not in (-1,0]$ such that $u(z) = Re(f(z))$. The point is that moving the contour you used for the Poisson kernel shows $f$ can be continued analytically along every curve $gamma subset 0 < |z| < 1$ and that when continued analytically along the closed curve $gamma : r e^it, t in [0,2pi]$ then $fmapsto f_gamma$ with $Re(f_gamma-f)=0$ thus $f_gamma-f$ is constant $=A in i BbbR$ and hence $f-fracA2ipi log z$ is analytic on $0< |z|< 1$
$endgroup$
– reuns
Apr 2 at 20:32
$begingroup$
Thanks for your explanations! This way it looks like the constant $A$ depends on the choice of the curve $gamma$. But you used $A$ to define $gin Hol(z)$. What am I missing here?
$endgroup$
– Yarin Luhmany
Apr 3 at 9:39
add a comment |
$begingroup$
I'm trying to solve the following problem:
Suppose that $u:mathbb D setminus0 to mathbb R$ is harmonic and that $lim_zto 0 u(z)=infty$.
Show that $u$ can be written as $$u(z)=beta ln |z| + v(z) $$ where $betain mathbb R setminus 0$ and $v:mathbb D to mathbb R $ is harmonic in $mathbb D$ (Here $mathbb D = z $).
I was also given the following hint:
Hint: Show that the residue of $(u_x - i u_y) $ is real.
I believe that the solution sould be related to Dirichlet's problem, and especially to Poisson's formula, but I don't understand how to use those to solve it.
Any help would will be appreciated!
complex-analysis harmonic-functions
asked Apr 1 at 18:15
Yarin LuhmanyYarin Luhmany
512
$endgroup$
I'm trying to solve the following problem:
Suppose that $u:mathbb D setminus0 to mathbb R$ is harmonic and that $lim_zto 0 u(z)=infty$.
Show that $u$ can be written as $$u(z)=beta ln |z| + v(z) $$ where $betain mathbb R setminus 0$ and $v:mathbb D to mathbb R $ is harmonic in $mathbb D$ (Here $mathbb D = z $).
I was also given the following hint:
Hint: Show that the residue of $(u_x - i u_y) $ is real.
I believe that the solution sould be related to Dirichlet's problem, and especially to Poisson's formula, but I don't understand how to use those to solve it.
Any help would will be appreciated!
complex-analysis harmonic-functions
complex-analysis harmonic-functions
asked Apr 1 at 18:15
Yarin LuhmanyYarin Luhmany
512
asked Apr 1 at 18:15
Yarin LuhmanyYarin Luhmany
512
asked Apr 1 at 18:15
Yarin LuhmanyYarin Luhmany
512
asked Apr 1 at 18:15
Yarin LuhmanyYarin Luhmany
512
asked Apr 1 at 18:15
Yarin LuhmanyYarin Luhmany
512
512
1
$begingroup$
On any simply connected open set $Usubset 0<$, $u(z) = Re(f(z))$ where $f$ is analytic on $U$, and the analytic continuation of $f$ is analytic over any curve $subset 0<$, and over closed curves, the continuation $f_gamma$ differs off $f$ by a purely imaginary constant. Let $A = lim_t to 0 f(e^2i pi (1-t))- f(e^2i pi t)$ which is purely imaginary. Then $g(z)=f(z)- A fraclog z2ipi$ is analytic on $0 < |z| < 1$.
$endgroup$
– reuns
Apr 1 at 23:18
1
$begingroup$
From there look at the 3 cases : $g$ has an essential singularity, a pole, a removable singularity, and show only in the latter case $lim_z to 0 |u(z)| = infty$.
$endgroup$
– reuns
Apr 1 at 23:18
$begingroup$
Thank you for your answer! Why can we use analytic continuation (for $f$)? Also I'd love an explantion for the definition of $A$ (we use $f(e^2pi i t)$, but $f$ is only defined on $Usubseteq 0<$....)
$endgroup$
– Yarin Luhmany
Apr 2 at 5:33
1
$begingroup$
From the Poisson kernel you can construct $f$ analytic on $0<|z|<1, z not in (-1,0]$ such that $u(z) = Re(f(z))$. The point is that moving the contour you used for the Poisson kernel shows $f$ can be continued analytically along every curve $gamma subset 0 < |z| < 1$ and that when continued analytically along the closed curve $gamma : r e^it, t in [0,2pi]$ then $fmapsto f_gamma$ with $Re(f_gamma-f)=0$ thus $f_gamma-f$ is constant $=A in i BbbR$ and hence $f-fracA2ipi log z$ is analytic on $0< |z|< 1$
$endgroup$
– reuns
Apr 2 at 20:32
$begingroup$
Thanks for your explanations! This way it looks like the constant $A$ depends on the choice of the curve $gamma$. But you used $A$ to define $gin Hol(z)$. What am I missing here?
$endgroup$
– Yarin Luhmany
Apr 3 at 9:39
add a comment |
1
$begingroup$
On any simply connected open set $Usubset 0<$, $u(z) = Re(f(z))$ where $f$ is analytic on $U$, and the analytic continuation of $f$ is analytic over any curve $subset 0<$, and over closed curves, the continuation $f_gamma$ differs off $f$ by a purely imaginary constant. Let $A = lim_t to 0 f(e^2i pi (1-t))- f(e^2i pi t)$ which is purely imaginary. Then $g(z)=f(z)- A fraclog z2ipi$ is analytic on $0 < |z| < 1$.
$endgroup$
– reuns
Apr 1 at 23:18
1
$begingroup$
From there look at the 3 cases : $g$ has an essential singularity, a pole, a removable singularity, and show only in the latter case $lim_z to 0 |u(z)| = infty$.
$endgroup$
– reuns
Apr 1 at 23:18
$begingroup$
Thank you for your answer! Why can we use analytic continuation (for $f$)? Also I'd love an explantion for the definition of $A$ (we use $f(e^2pi i t)$, but $f$ is only defined on $Usubseteq 0<$....)
$endgroup$
– Yarin Luhmany
Apr 2 at 5:33
1
$begingroup$
From the Poisson kernel you can construct $f$ analytic on $0<|z|<1, z not in (-1,0]$ such that $u(z) = Re(f(z))$. The point is that moving the contour you used for the Poisson kernel shows $f$ can be continued analytically along every curve $gamma subset 0 < |z| < 1$ and that when continued analytically along the closed curve $gamma : r e^it, t in [0,2pi]$ then $fmapsto f_gamma$ with $Re(f_gamma-f)=0$ thus $f_gamma-f$ is constant $=A in i BbbR$ and hence $f-fracA2ipi log z$ is analytic on $0< |z|< 1$
$endgroup$
– reuns
Apr 2 at 20:32
$begingroup$
Thanks for your explanations! This way it looks like the constant $A$ depends on the choice of the curve $gamma$. But you used $A$ to define $gin Hol(z)$. What am I missing here?
$endgroup$
– Yarin Luhmany
Apr 3 at 9:39
1
1
$begingroup$
On any simply connected open set $Usubset 0<$, $u(z) = Re(f(z))$ where $f$ is analytic on $U$, and the analytic continuation of $f$ is analytic over any curve $subset 0<$, and over closed curves, the continuation $f_gamma$ differs off $f$ by a purely imaginary constant. Let $A = lim_t to 0 f(e^2i pi (1-t))- f(e^2i pi t)$ which is purely imaginary. Then $g(z)=f(z)- A fraclog z2ipi$ is analytic on $0 < |z| < 1$.
$endgroup$
– reuns
Apr 1 at 23:18
$begingroup$
On any simply connected open set $Usubset 0<$, $u(z) = Re(f(z))$ where $f$ is analytic on $U$, and the analytic continuation of $f$ is analytic over any curve $subset 0<$, and over closed curves, the continuation $f_gamma$ differs off $f$ by a purely imaginary constant. Let $A = lim_t to 0 f(e^2i pi (1-t))- f(e^2i pi t)$ which is purely imaginary. Then $g(z)=f(z)- A fraclog z2ipi$ is analytic on $0 < |z| < 1$.
$endgroup$
– reuns
Apr 1 at 23:18
1
1
$begingroup$
From there look at the 3 cases : $g$ has an essential singularity, a pole, a removable singularity, and show only in the latter case $lim_z to 0 |u(z)| = infty$.
$endgroup$
– reuns
Apr 1 at 23:18
$begingroup$
From there look at the 3 cases : $g$ has an essential singularity, a pole, a removable singularity, and show only in the latter case $lim_z to 0 |u(z)| = infty$.
$endgroup$
– reuns
Apr 1 at 23:18
$begingroup$
Thank you for your answer! Why can we use analytic continuation (for $f$)? Also I'd love an explantion for the definition of $A$ (we use $f(e^2pi i t)$, but $f$ is only defined on $Usubseteq 0<$....)
$endgroup$
– Yarin Luhmany
Apr 2 at 5:33
$begingroup$
Thank you for your answer! Why can we use analytic continuation (for $f$)? Also I'd love an explantion for the definition of $A$ (we use $f(e^2pi i t)$, but $f$ is only defined on $Usubseteq 0<$....)
$endgroup$
– Yarin Luhmany
Apr 2 at 5:33
1
1
$begingroup$
From the Poisson kernel you can construct $f$ analytic on $0<|z|<1, z not in (-1,0]$ such that $u(z) = Re(f(z))$. The point is that moving the contour you used for the Poisson kernel shows $f$ can be continued analytically along every curve $gamma subset 0 < |z| < 1$ and that when continued analytically along the closed curve $gamma : r e^it, t in [0,2pi]$ then $fmapsto f_gamma$ with $Re(f_gamma-f)=0$ thus $f_gamma-f$ is constant $=A in i BbbR$ and hence $f-fracA2ipi log z$ is analytic on $0< |z|< 1$
$endgroup$
– reuns
Apr 2 at 20:32
$begingroup$
From the Poisson kernel you can construct $f$ analytic on $0<|z|<1, z not in (-1,0]$ such that $u(z) = Re(f(z))$. The point is that moving the contour you used for the Poisson kernel shows $f$ can be continued analytically along every curve $gamma subset 0 < |z| < 1$ and that when continued analytically along the closed curve $gamma : r e^it, t in [0,2pi]$ then $fmapsto f_gamma$ with $Re(f_gamma-f)=0$ thus $f_gamma-f$ is constant $=A in i BbbR$ and hence $f-fracA2ipi log z$ is analytic on $0< |z|< 1$
$endgroup$
– reuns
Apr 2 at 20:32
$begingroup$
Thanks for your explanations! This way it looks like the constant $A$ depends on the choice of the curve $gamma$. But you used $A$ to define $gin Hol(z)$. What am I missing here?
$endgroup$
– Yarin Luhmany
Apr 3 at 9:39
$begingroup$
Thanks for your explanations! This way it looks like the constant $A$ depends on the choice of the curve $gamma$. But you used $A$ to define $gin Hol(z)$. What am I missing here?
$endgroup$
– Yarin Luhmany
Apr 3 at 9:39
add a comment |
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$begingroup$
On any simply connected open set $Usubset 0<$, $u(z) = Re(f(z))$ where $f$ is analytic on $U$, and the analytic continuation of $f$ is analytic over any curve $subset 0<$, and over closed curves, the continuation $f_gamma$ differs off $f$ by a purely imaginary constant. Let $A = lim_t to 0 f(e^2i pi (1-t))- f(e^2i pi t)$ which is purely imaginary. Then $g(z)=f(z)- A fraclog z2ipi$ is analytic on $0 < |z| < 1$.
$endgroup$
– reuns
Apr 1 at 23:18
1
$begingroup$
From there look at the 3 cases : $g$ has an essential singularity, a pole, a removable singularity, and show only in the latter case $lim_z to 0 |u(z)| = infty$.
$endgroup$
– reuns
Apr 1 at 23:18
$begingroup$
Thank you for your answer! Why can we use analytic continuation (for $f$)? Also I'd love an explantion for the definition of $A$ (we use $f(e^2pi i t)$, but $f$ is only defined on $Usubseteq 0<$....)
$endgroup$
– Yarin Luhmany
Apr 2 at 5:33
1
$begingroup$
From the Poisson kernel you can construct $f$ analytic on $0<|z|<1, z not in (-1,0]$ such that $u(z) = Re(f(z))$. The point is that moving the contour you used for the Poisson kernel shows $f$ can be continued analytically along every curve $gamma subset 0 < |z| < 1$ and that when continued analytically along the closed curve $gamma : r e^it, t in [0,2pi]$ then $fmapsto f_gamma$ with $Re(f_gamma-f)=0$ thus $f_gamma-f$ is constant $=A in i BbbR$ and hence $f-fracA2ipi log z$ is analytic on $0< |z|< 1$
$endgroup$
– reuns
Apr 2 at 20:32
$begingroup$
Thanks for your explanations! This way it looks like the constant $A$ depends on the choice of the curve $gamma$. But you used $A$ to define $gin Hol(z)$. What am I missing here?
$endgroup$
– Yarin Luhmany
Apr 3 at 9:39