Using conformal maps to solve the Dirichlet problem on $U = {z : text{Im}z geq 0 }$
I'm trying to solve a question which asks me to solve the Dirichlet problem on $U = {z : text{Im}z geq 0 }$ on the condition that (where we let $z = x+iy$) $u(x,0) = 0 $ when $ |x| >1 $ and $ u(x,0) = 1 $ when $ |x| < 1$.
I've already solved the problem for $U = {z : text{Im}z geq 0 }$ with $u(x,0) = 0 $ when $x > 0 $ and $ u(x,0) = 1 $ when $ x< 0$, and I presume there is some way to use a conformal map sending the positive left real axis to the real numbers with $|x| > 1$ and the negative to those with $|x|<1$, but I'm struggling to see how.
Another result I have from an earlier question is the solution on $U = {x+iy : 0 leq y leq 1 }$, $u(x,0) = 0$, $u(x,1)=1$ as $u(x,y) = y$, but I'm not sure that would be so useful.
If anyone knows any map which would relate the two I'd really appreciate your help, or if you know of another way to do this using the other result and conformal maps, I'd be really interested to hear that too.
complex-analysis complex-numbers conformal-geometry mobius-transformation
asked 2 days ago
xujxuj
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I'm trying to solve a question which asks me to solve the Dirichlet problem on $U = {z : text{Im}z geq 0 }$ on the condition that (where we let $z = x+iy$) $u(x,0) = 0 $ when $ |x| >1 $ and $ u(x,0) = 1 $ when $ |x| < 1$.
I've already solved the problem for $U = {z : text{Im}z geq 0 }$ with $u(x,0) = 0 $ when $x > 0 $ and $ u(x,0) = 1 $ when $ x< 0$, and I presume there is some way to use a conformal map sending the positive left real axis to the real numbers with $|x| > 1$ and the negative to those with $|x|<1$, but I'm struggling to see how.
Another result I have from an earlier question is the solution on $U = {x+iy : 0 leq y leq 1 }$, $u(x,0) = 0$, $u(x,1)=1$ as $u(x,y) = y$, but I'm not sure that would be so useful.
If anyone knows any map which would relate the two I'd really appreciate your help, or if you know of another way to do this using the other result and conformal maps, I'd be really interested to hear that too.
complex-analysis complex-numbers conformal-geometry mobius-transformation
asked 2 days ago
xujxuj
82
New contributor
xuj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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I'm trying to solve a question which asks me to solve the Dirichlet problem on $U = {z : text{Im}z geq 0 }$ on the condition that (where we let $z = x+iy$) $u(x,0) = 0 $ when $ |x| >1 $ and $ u(x,0) = 1 $ when $ |x| < 1$.
I've already solved the problem for $U = {z : text{Im}z geq 0 }$ with $u(x,0) = 0 $ when $x > 0 $ and $ u(x,0) = 1 $ when $ x< 0$, and I presume there is some way to use a conformal map sending the positive left real axis to the real numbers with $|x| > 1$ and the negative to those with $|x|<1$, but I'm struggling to see how.
Another result I have from an earlier question is the solution on $U = {x+iy : 0 leq y leq 1 }$, $u(x,0) = 0$, $u(x,1)=1$ as $u(x,y) = y$, but I'm not sure that would be so useful.
If anyone knows any map which would relate the two I'd really appreciate your help, or if you know of another way to do this using the other result and conformal maps, I'd be really interested to hear that too.
complex-analysis complex-numbers conformal-geometry mobius-transformation
asked 2 days ago
xujxuj
82
New contributor
xuj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm trying to solve a question which asks me to solve the Dirichlet problem on $U = {z : text{Im}z geq 0 }$ on the condition that (where we let $z = x+iy$) $u(x,0) = 0 $ when $ |x| >1 $ and $ u(x,0) = 1 $ when $ |x| < 1$.
I've already solved the problem for $U = {z : text{Im}z geq 0 }$ with $u(x,0) = 0 $ when $x > 0 $ and $ u(x,0) = 1 $ when $ x< 0$, and I presume there is some way to use a conformal map sending the positive left real axis to the real numbers with $|x| > 1$ and the negative to those with $|x|<1$, but I'm struggling to see how.
Another result I have from an earlier question is the solution on $U = {x+iy : 0 leq y leq 1 }$, $u(x,0) = 0$, $u(x,1)=1$ as $u(x,y) = y$, but I'm not sure that would be so useful.
If anyone knows any map which would relate the two I'd really appreciate your help, or if you know of another way to do this using the other result and conformal maps, I'd be really interested to hear that too.
complex-analysis complex-numbers conformal-geometry mobius-transformation
complex-analysis complex-numbers conformal-geometry mobius-transformation
asked 2 days ago
xujxuj
82
New contributor
xuj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
xujxuj
82
New contributor
xuj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
xujxuj
82
New contributor
xuj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
xujxuj
82
asked 2 days ago
xujxuj
82
82
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xuj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
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1 Answer
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All conformal maps of the upper half plane are Möbius transformations of the
form
$$
T(z) = frac{az+b}{cz+d}
$$
with $a, b, c, d in Bbb R$, $ad-bc > 0$. We need that $(-infty, 0)$ is mapped to $(-1, 1)$, and since conformal maps preserve the orientation,
$$
T(infty) = -1 , , quad T(0) = 1
$$
must hold. Now it should not be too difficult to find that
$$
T(z) = frac{1+z}{1-z}
$$
satisfies all the needs, and allows to transform a solution of your second Dirichlet problem to a solution of your first problem.
answered 2 days ago
Martin RMartin R
27.3k33254
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
All conformal maps of the upper half plane are Möbius transformations of the
form
$$
T(z) = frac{az+b}{cz+d}
$$
with $a, b, c, d in Bbb R$, $ad-bc > 0$. We need that $(-infty, 0)$ is mapped to $(-1, 1)$, and since conformal maps preserve the orientation,
$$
T(infty) = -1 , , quad T(0) = 1
$$
must hold. Now it should not be too difficult to find that
$$
T(z) = frac{1+z}{1-z}
$$
satisfies all the needs, and allows to transform a solution of your second Dirichlet problem to a solution of your first problem.
answered 2 days ago
Martin RMartin R
27.3k33254
add a comment |
All conformal maps of the upper half plane are Möbius transformations of the
form
$$
T(z) = frac{az+b}{cz+d}
$$
with $a, b, c, d in Bbb R$, $ad-bc > 0$. We need that $(-infty, 0)$ is mapped to $(-1, 1)$, and since conformal maps preserve the orientation,
$$
T(infty) = -1 , , quad T(0) = 1
$$
must hold. Now it should not be too difficult to find that
$$
T(z) = frac{1+z}{1-z}
$$
satisfies all the needs, and allows to transform a solution of your second Dirichlet problem to a solution of your first problem.
answered 2 days ago
Martin RMartin R
27.3k33254
add a comment |
All conformal maps of the upper half plane are Möbius transformations of the
form
$$
T(z) = frac{az+b}{cz+d}
$$
with $a, b, c, d in Bbb R$, $ad-bc > 0$. We need that $(-infty, 0)$ is mapped to $(-1, 1)$, and since conformal maps preserve the orientation,
$$
T(infty) = -1 , , quad T(0) = 1
$$
must hold. Now it should not be too difficult to find that
$$
T(z) = frac{1+z}{1-z}
$$
satisfies all the needs, and allows to transform a solution of your second Dirichlet problem to a solution of your first problem.
answered 2 days ago
Martin RMartin R
27.3k33254
All conformal maps of the upper half plane are Möbius transformations of the
form
$$
T(z) = frac{az+b}{cz+d}
$$
with $a, b, c, d in Bbb R$, $ad-bc > 0$. We need that $(-infty, 0)$ is mapped to $(-1, 1)$, and since conformal maps preserve the orientation,
$$
T(infty) = -1 , , quad T(0) = 1
$$
must hold. Now it should not be too difficult to find that
$$
T(z) = frac{1+z}{1-z}
$$
satisfies all the needs, and allows to transform a solution of your second Dirichlet problem to a solution of your first problem.
answered 2 days ago
Martin RMartin R
27.3k33254
edited 2 days ago
answered 2 days ago
Martin RMartin R
27.3k33254
answered 2 days ago
Martin RMartin R
27.3k33254
answered 2 days ago
Martin RMartin R
27.3k33254
27.3k33254
add a comment |
add a comment |
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