Solving the Dirichlet problem on $ U = {z: text{Im} z geq 0}$ using a conformal map
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I'm trying to solve the Dirichlet problem on $ U = {z: text{Im} z geq 0}$ with the conditions $u(x,0) = 0$ when $x>0$, $u(x,0)=1$ when $x<0$.
To do so, I'm supposed to use conformal maps along with the fact that the solution to the Dirichlet problem on $U = {x+ i y : 0 leq y leq 1} $ with $ u(x,0) = 0, u(x,1) = 1$ is just $u(x,y)=y$.
The map $e^{pi z}$ takes ${x+ i y : 0 leq y leq 1}$ to ${z: text{Im} z geq 0}setminus{0}$, and so taking the inverse of this - $frac{1}{pi}log z$ and then taking the imaginary part of this (i.e. applying $u(x,y) = y$) gives me $u(z) = frac{1}{pi}arg z$, which gives the required result.
The problem I have here is that this isn't defined at $z=0$, and I'm not sure what other solution would work.
Does anyone have any other ideas that might help?
complex-analysis complex-numbers conformal-geometry mobius-transformation
asked Jan 5 at 0:00
jdoejdoe
11
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add a comment |
$begingroup$
I'm trying to solve the Dirichlet problem on $ U = {z: text{Im} z geq 0}$ with the conditions $u(x,0) = 0$ when $x>0$, $u(x,0)=1$ when $x<0$.
To do so, I'm supposed to use conformal maps along with the fact that the solution to the Dirichlet problem on $U = {x+ i y : 0 leq y leq 1} $ with $ u(x,0) = 0, u(x,1) = 1$ is just $u(x,y)=y$.
The map $e^{pi z}$ takes ${x+ i y : 0 leq y leq 1}$ to ${z: text{Im} z geq 0}setminus{0}$, and so taking the inverse of this - $frac{1}{pi}log z$ and then taking the imaginary part of this (i.e. applying $u(x,y) = y$) gives me $u(z) = frac{1}{pi}arg z$, which gives the required result.
The problem I have here is that this isn't defined at $z=0$, and I'm not sure what other solution would work.
Does anyone have any other ideas that might help?
complex-analysis complex-numbers conformal-geometry mobius-transformation
asked Jan 5 at 0:00
jdoejdoe
11
$endgroup$
$begingroup$
This is not really a problem, since your solution is necessarily discontinuous at $z=0$, and harmonicity is only required in the open upper halfplane. I.e., your solution is correct.
$endgroup$
– Lukas Geyer
Jan 5 at 6:41
add a comment |
$begingroup$
I'm trying to solve the Dirichlet problem on $ U = {z: text{Im} z geq 0}$ with the conditions $u(x,0) = 0$ when $x>0$, $u(x,0)=1$ when $x<0$.
To do so, I'm supposed to use conformal maps along with the fact that the solution to the Dirichlet problem on $U = {x+ i y : 0 leq y leq 1} $ with $ u(x,0) = 0, u(x,1) = 1$ is just $u(x,y)=y$.
The map $e^{pi z}$ takes ${x+ i y : 0 leq y leq 1}$ to ${z: text{Im} z geq 0}setminus{0}$, and so taking the inverse of this - $frac{1}{pi}log z$ and then taking the imaginary part of this (i.e. applying $u(x,y) = y$) gives me $u(z) = frac{1}{pi}arg z$, which gives the required result.
The problem I have here is that this isn't defined at $z=0$, and I'm not sure what other solution would work.
Does anyone have any other ideas that might help?
complex-analysis complex-numbers conformal-geometry mobius-transformation
asked Jan 5 at 0:00
jdoejdoe
11
$endgroup$
I'm trying to solve the Dirichlet problem on $ U = {z: text{Im} z geq 0}$ with the conditions $u(x,0) = 0$ when $x>0$, $u(x,0)=1$ when $x<0$.
To do so, I'm supposed to use conformal maps along with the fact that the solution to the Dirichlet problem on $U = {x+ i y : 0 leq y leq 1} $ with $ u(x,0) = 0, u(x,1) = 1$ is just $u(x,y)=y$.
The map $e^{pi z}$ takes ${x+ i y : 0 leq y leq 1}$ to ${z: text{Im} z geq 0}setminus{0}$, and so taking the inverse of this - $frac{1}{pi}log z$ and then taking the imaginary part of this (i.e. applying $u(x,y) = y$) gives me $u(z) = frac{1}{pi}arg z$, which gives the required result.
The problem I have here is that this isn't defined at $z=0$, and I'm not sure what other solution would work.
Does anyone have any other ideas that might help?
complex-analysis complex-numbers conformal-geometry mobius-transformation
complex-analysis complex-numbers conformal-geometry mobius-transformation
asked Jan 5 at 0:00
jdoejdoe
11
asked Jan 5 at 0:00
jdoejdoe
11
edited Jan 5 at 0:16
jdoe
asked Jan 5 at 0:00
jdoejdoe
11
asked Jan 5 at 0:00
jdoejdoe
11
asked Jan 5 at 0:00
jdoejdoe
11
11
$begingroup$
This is not really a problem, since your solution is necessarily discontinuous at $z=0$, and harmonicity is only required in the open upper halfplane. I.e., your solution is correct.
$endgroup$
– Lukas Geyer
Jan 5 at 6:41
add a comment |
$begingroup$
This is not really a problem, since your solution is necessarily discontinuous at $z=0$, and harmonicity is only required in the open upper halfplane. I.e., your solution is correct.
$endgroup$
– Lukas Geyer
Jan 5 at 6:41
$begingroup$
This is not really a problem, since your solution is necessarily discontinuous at $z=0$, and harmonicity is only required in the open upper halfplane. I.e., your solution is correct.
$endgroup$
– Lukas Geyer
Jan 5 at 6:41
$begingroup$
This is not really a problem, since your solution is necessarily discontinuous at $z=0$, and harmonicity is only required in the open upper halfplane. I.e., your solution is correct.
$endgroup$
– Lukas Geyer
Jan 5 at 6:41
add a comment |
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This is not really a problem, since your solution is necessarily discontinuous at $z=0$, and harmonicity is only required in the open upper halfplane. I.e., your solution is correct.
$endgroup$
– Lukas Geyer
Jan 5 at 6:41