Using conformal maps to solve the Dirichlet problem on $U = {z : text{Im}z geq 0 }$
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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 Jan 5 at 19:02
xujxuj
82
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add a comment |
$begingroup$
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 Jan 5 at 19:02
xujxuj
82
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 19:02
xujxuj
82
$endgroup$
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 Jan 5 at 19:02
xujxuj
82
asked Jan 5 at 19:02
xujxuj
82
asked Jan 5 at 19:02
xujxuj
82
asked Jan 5 at 19:02
xujxuj
82
asked Jan 5 at 19:02
xujxuj
82
82
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
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 Jan 5 at 19:21
Martin RMartin R
27.5k33255
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 Jan 5 at 19:21
Martin RMartin R
27.5k33255
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 19:21
Martin RMartin R
27.5k33255
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 19:21
Martin RMartin R
27.5k33255
$endgroup$
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 Jan 5 at 19:21
Martin RMartin R
27.5k33255
edited Jan 5 at 19:40
answered Jan 5 at 19:21
Martin RMartin R
27.5k33255
answered Jan 5 at 19:21
Martin RMartin R
27.5k33255
answered Jan 5 at 19:21
Martin RMartin R
27.5k33255
27.5k33255
add a comment |
add a comment |
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