Biholomorphism between lower-half-disk and upper-half-plane, with two conditions.
$begingroup$
I would like to find a holomorphic map that transforms the lower half unit-disk into the upper half-plane, with two constraints:
- the image of the boundary [-1,1] of the half-disk is $infty$
- the image of the boundary point -i is 0.
Thank you if someone has ideas.
Edit: (some precisions on background)
My question comes from analysis of elliptic PDEs.
I know an explicit formula for a harmonic function $u$ in the upper half-plane $mathbb{R} times mathbb{R}_+^star$, that satisfies a certain Neumann boundary condition on the boundary $mathbb{R} times { 0 }$. The boundary condition is precisely $frac{partial u}{partialnu} = -frac{1}{2}sin(2(u-g))$ where $g$ is given by $nu=ie^{ig}$, $nu$ being the outer unit normal on the boundary of the domain.
Now, I would like to find an explicit formula for a harmonic function in the lower half-disk, that satisfies the same Neumann boundary condition as before, on a boundary that remains to find (but it is not the question here).
My strategy is to transform the upper half-plane into the lower half-disk using a biholomorphism $phi$, because I will then be sure that the composition $u circ phi^{-1}$ is harmonic in the lower half-disk (as a composition of a harmonic function and a holomorphic function).
If I represent the vectors $(cos u,sin u)$ in the upper half-plane, the point $0$ is "repulsive". As I want additionally to keep the "behaviour" of these vectors for the composition $u circ phi^{-1}$, I ask two conditions for $phi^{-1}$: $phi^{-1}(-i)=0$ and $phi^{-1}(x)=infty$ for $xin [-1,1]$.
complex-analysis holomorphic-functions
asked Jan 25 at 16:20
lofralofra
63
$endgroup$
add a comment |
$begingroup$
I would like to find a holomorphic map that transforms the lower half unit-disk into the upper half-plane, with two constraints:
- the image of the boundary [-1,1] of the half-disk is $infty$
- the image of the boundary point -i is 0.
Thank you if someone has ideas.
Edit: (some precisions on background)
My question comes from analysis of elliptic PDEs.
I know an explicit formula for a harmonic function $u$ in the upper half-plane $mathbb{R} times mathbb{R}_+^star$, that satisfies a certain Neumann boundary condition on the boundary $mathbb{R} times { 0 }$. The boundary condition is precisely $frac{partial u}{partialnu} = -frac{1}{2}sin(2(u-g))$ where $g$ is given by $nu=ie^{ig}$, $nu$ being the outer unit normal on the boundary of the domain.
Now, I would like to find an explicit formula for a harmonic function in the lower half-disk, that satisfies the same Neumann boundary condition as before, on a boundary that remains to find (but it is not the question here).
My strategy is to transform the upper half-plane into the lower half-disk using a biholomorphism $phi$, because I will then be sure that the composition $u circ phi^{-1}$ is harmonic in the lower half-disk (as a composition of a harmonic function and a holomorphic function).
If I represent the vectors $(cos u,sin u)$ in the upper half-plane, the point $0$ is "repulsive". As I want additionally to keep the "behaviour" of these vectors for the composition $u circ phi^{-1}$, I ask two conditions for $phi^{-1}$: $phi^{-1}(-i)=0$ and $phi^{-1}(x)=infty$ for $xin [-1,1]$.
complex-analysis holomorphic-functions
asked Jan 25 at 16:20
lofralofra
63
$endgroup$
$begingroup$
Hint: start with the following mobius map f(z)=(z-1)/(z+1).. later on you'll might want use g(z)=z^2
$endgroup$
– zokomoko
Jan 25 at 18:14
$begingroup$
Thank you @zokomoko. If I follow your hint, I think you want to say that your map f transforms the lower half-disk into the quadrant ${ x,y<0 }$, then I can multiply by $i^2$ to obtain the first quadrant, and then take the square to obtain the upper half-plane. So the map would be $z mapsto ((z-1)/(z+1))^2$, but my constraints are not satisfied... The image of $[-1,1]$ is the positive real axis and the image of $-i$ is $-1$, if I am not wrong.
$endgroup$
– lofra
Jan 26 at 13:03
$begingroup$
Any conformal map between Jordan domains (on the sphere) extends to a homeomorphism of the closures, so it is not possible that your map takes a whole interval to just the one point $infty$.
$endgroup$
– Lukas Geyer
Jan 27 at 16:47
add a comment |
$begingroup$
I would like to find a holomorphic map that transforms the lower half unit-disk into the upper half-plane, with two constraints:
- the image of the boundary [-1,1] of the half-disk is $infty$
- the image of the boundary point -i is 0.
Thank you if someone has ideas.
Edit: (some precisions on background)
My question comes from analysis of elliptic PDEs.
I know an explicit formula for a harmonic function $u$ in the upper half-plane $mathbb{R} times mathbb{R}_+^star$, that satisfies a certain Neumann boundary condition on the boundary $mathbb{R} times { 0 }$. The boundary condition is precisely $frac{partial u}{partialnu} = -frac{1}{2}sin(2(u-g))$ where $g$ is given by $nu=ie^{ig}$, $nu$ being the outer unit normal on the boundary of the domain.
Now, I would like to find an explicit formula for a harmonic function in the lower half-disk, that satisfies the same Neumann boundary condition as before, on a boundary that remains to find (but it is not the question here).
My strategy is to transform the upper half-plane into the lower half-disk using a biholomorphism $phi$, because I will then be sure that the composition $u circ phi^{-1}$ is harmonic in the lower half-disk (as a composition of a harmonic function and a holomorphic function).
If I represent the vectors $(cos u,sin u)$ in the upper half-plane, the point $0$ is "repulsive". As I want additionally to keep the "behaviour" of these vectors for the composition $u circ phi^{-1}$, I ask two conditions for $phi^{-1}$: $phi^{-1}(-i)=0$ and $phi^{-1}(x)=infty$ for $xin [-1,1]$.
complex-analysis holomorphic-functions
asked Jan 25 at 16:20
lofralofra
63
$endgroup$
I would like to find a holomorphic map that transforms the lower half unit-disk into the upper half-plane, with two constraints:
- the image of the boundary [-1,1] of the half-disk is $infty$
- the image of the boundary point -i is 0.
Thank you if someone has ideas.
Edit: (some precisions on background)
My question comes from analysis of elliptic PDEs.
I know an explicit formula for a harmonic function $u$ in the upper half-plane $mathbb{R} times mathbb{R}_+^star$, that satisfies a certain Neumann boundary condition on the boundary $mathbb{R} times { 0 }$. The boundary condition is precisely $frac{partial u}{partialnu} = -frac{1}{2}sin(2(u-g))$ where $g$ is given by $nu=ie^{ig}$, $nu$ being the outer unit normal on the boundary of the domain.
Now, I would like to find an explicit formula for a harmonic function in the lower half-disk, that satisfies the same Neumann boundary condition as before, on a boundary that remains to find (but it is not the question here).
My strategy is to transform the upper half-plane into the lower half-disk using a biholomorphism $phi$, because I will then be sure that the composition $u circ phi^{-1}$ is harmonic in the lower half-disk (as a composition of a harmonic function and a holomorphic function).
If I represent the vectors $(cos u,sin u)$ in the upper half-plane, the point $0$ is "repulsive". As I want additionally to keep the "behaviour" of these vectors for the composition $u circ phi^{-1}$, I ask two conditions for $phi^{-1}$: $phi^{-1}(-i)=0$ and $phi^{-1}(x)=infty$ for $xin [-1,1]$.
complex-analysis holomorphic-functions
complex-analysis holomorphic-functions
asked Jan 25 at 16:20
lofralofra
63
asked Jan 25 at 16:20
lofralofra
63
edited Jan 27 at 11:11
lofra
asked Jan 25 at 16:20
lofralofra
63
asked Jan 25 at 16:20
lofralofra
63
asked Jan 25 at 16:20
lofralofra
63
63
$begingroup$
Hint: start with the following mobius map f(z)=(z-1)/(z+1).. later on you'll might want use g(z)=z^2
$endgroup$
– zokomoko
Jan 25 at 18:14
$begingroup$
Thank you @zokomoko. If I follow your hint, I think you want to say that your map f transforms the lower half-disk into the quadrant ${ x,y<0 }$, then I can multiply by $i^2$ to obtain the first quadrant, and then take the square to obtain the upper half-plane. So the map would be $z mapsto ((z-1)/(z+1))^2$, but my constraints are not satisfied... The image of $[-1,1]$ is the positive real axis and the image of $-i$ is $-1$, if I am not wrong.
$endgroup$
– lofra
Jan 26 at 13:03
$begingroup$
Any conformal map between Jordan domains (on the sphere) extends to a homeomorphism of the closures, so it is not possible that your map takes a whole interval to just the one point $infty$.
$endgroup$
– Lukas Geyer
Jan 27 at 16:47
add a comment |
$begingroup$
Hint: start with the following mobius map f(z)=(z-1)/(z+1).. later on you'll might want use g(z)=z^2
$endgroup$
– zokomoko
Jan 25 at 18:14
$begingroup$
Thank you @zokomoko. If I follow your hint, I think you want to say that your map f transforms the lower half-disk into the quadrant ${ x,y<0 }$, then I can multiply by $i^2$ to obtain the first quadrant, and then take the square to obtain the upper half-plane. So the map would be $z mapsto ((z-1)/(z+1))^2$, but my constraints are not satisfied... The image of $[-1,1]$ is the positive real axis and the image of $-i$ is $-1$, if I am not wrong.
$endgroup$
– lofra
Jan 26 at 13:03
$begingroup$
Any conformal map between Jordan domains (on the sphere) extends to a homeomorphism of the closures, so it is not possible that your map takes a whole interval to just the one point $infty$.
$endgroup$
– Lukas Geyer
Jan 27 at 16:47
$begingroup$
Hint: start with the following mobius map f(z)=(z-1)/(z+1).. later on you'll might want use g(z)=z^2
$endgroup$
– zokomoko
Jan 25 at 18:14
$begingroup$
Hint: start with the following mobius map f(z)=(z-1)/(z+1).. later on you'll might want use g(z)=z^2
$endgroup$
– zokomoko
Jan 25 at 18:14
$begingroup$
Thank you @zokomoko. If I follow your hint, I think you want to say that your map f transforms the lower half-disk into the quadrant ${ x,y<0 }$, then I can multiply by $i^2$ to obtain the first quadrant, and then take the square to obtain the upper half-plane. So the map would be $z mapsto ((z-1)/(z+1))^2$, but my constraints are not satisfied... The image of $[-1,1]$ is the positive real axis and the image of $-i$ is $-1$, if I am not wrong.
$endgroup$
– lofra
Jan 26 at 13:03
$begingroup$
Thank you @zokomoko. If I follow your hint, I think you want to say that your map f transforms the lower half-disk into the quadrant ${ x,y<0 }$, then I can multiply by $i^2$ to obtain the first quadrant, and then take the square to obtain the upper half-plane. So the map would be $z mapsto ((z-1)/(z+1))^2$, but my constraints are not satisfied... The image of $[-1,1]$ is the positive real axis and the image of $-i$ is $-1$, if I am not wrong.
$endgroup$
– lofra
Jan 26 at 13:03
$begingroup$
Any conformal map between Jordan domains (on the sphere) extends to a homeomorphism of the closures, so it is not possible that your map takes a whole interval to just the one point $infty$.
$endgroup$
– Lukas Geyer
Jan 27 at 16:47
$begingroup$
Any conformal map between Jordan domains (on the sphere) extends to a homeomorphism of the closures, so it is not possible that your map takes a whole interval to just the one point $infty$.
$endgroup$
– Lukas Geyer
Jan 27 at 16:47
add a comment |
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$begingroup$
Hint: start with the following mobius map f(z)=(z-1)/(z+1).. later on you'll might want use g(z)=z^2
$endgroup$
– zokomoko
Jan 25 at 18:14
$begingroup$
Thank you @zokomoko. If I follow your hint, I think you want to say that your map f transforms the lower half-disk into the quadrant ${ x,y<0 }$, then I can multiply by $i^2$ to obtain the first quadrant, and then take the square to obtain the upper half-plane. So the map would be $z mapsto ((z-1)/(z+1))^2$, but my constraints are not satisfied... The image of $[-1,1]$ is the positive real axis and the image of $-i$ is $-1$, if I am not wrong.
$endgroup$
– lofra
Jan 26 at 13:03
$begingroup$
Any conformal map between Jordan domains (on the sphere) extends to a homeomorphism of the closures, so it is not possible that your map takes a whole interval to just the one point $infty$.
$endgroup$
– Lukas Geyer
Jan 27 at 16:47