Solving the Dirichlet Problem for an infinite strip
$begingroup$
I have been looking into the Dirichlet problem and conformal mappings, but am unsure as to how to find a solution $u(x, y)$ for the Dirichlet problem given this information:
The region is $U = { x+iy : 0 leq y leq 1 }$, and the boundary conditions are $u(x, 0) = 0$, $u(x, 1) = 1$.
I see this as an infinite strip in the upper plane. I am unsure then how to find a solution to the Dirichlet problem. I tried inscribing a disk within the strip and then using standard results for a disk, but ran into problems as the radius of the inscribed ball would have to be $lt 1$.
Does anyone have any thoughts or ideas about different functions I could try that would work?
complex-analysis harmonic-functions conformal-geometry
edited Jan 8 at 18:52
user35443
201213
asked Jan 8 at 12:33
jessg12345jessg12345
64
$endgroup$
add a comment |
$begingroup$
I have been looking into the Dirichlet problem and conformal mappings, but am unsure as to how to find a solution $u(x, y)$ for the Dirichlet problem given this information:
The region is $U = { x+iy : 0 leq y leq 1 }$, and the boundary conditions are $u(x, 0) = 0$, $u(x, 1) = 1$.
I see this as an infinite strip in the upper plane. I am unsure then how to find a solution to the Dirichlet problem. I tried inscribing a disk within the strip and then using standard results for a disk, but ran into problems as the radius of the inscribed ball would have to be $lt 1$.
Does anyone have any thoughts or ideas about different functions I could try that would work?
complex-analysis harmonic-functions conformal-geometry
edited Jan 8 at 18:52
user35443
201213
asked Jan 8 at 12:33
jessg12345jessg12345
64
$endgroup$
add a comment |
$begingroup$
I have been looking into the Dirichlet problem and conformal mappings, but am unsure as to how to find a solution $u(x, y)$ for the Dirichlet problem given this information:
The region is $U = { x+iy : 0 leq y leq 1 }$, and the boundary conditions are $u(x, 0) = 0$, $u(x, 1) = 1$.
I see this as an infinite strip in the upper plane. I am unsure then how to find a solution to the Dirichlet problem. I tried inscribing a disk within the strip and then using standard results for a disk, but ran into problems as the radius of the inscribed ball would have to be $lt 1$.
Does anyone have any thoughts or ideas about different functions I could try that would work?
complex-analysis harmonic-functions conformal-geometry
edited Jan 8 at 18:52
user35443
201213
asked Jan 8 at 12:33
jessg12345jessg12345
64
$endgroup$
I have been looking into the Dirichlet problem and conformal mappings, but am unsure as to how to find a solution $u(x, y)$ for the Dirichlet problem given this information:
The region is $U = { x+iy : 0 leq y leq 1 }$, and the boundary conditions are $u(x, 0) = 0$, $u(x, 1) = 1$.
I see this as an infinite strip in the upper plane. I am unsure then how to find a solution to the Dirichlet problem. I tried inscribing a disk within the strip and then using standard results for a disk, but ran into problems as the radius of the inscribed ball would have to be $lt 1$.
Does anyone have any thoughts or ideas about different functions I could try that would work?
complex-analysis harmonic-functions conformal-geometry
complex-analysis harmonic-functions conformal-geometry
edited Jan 8 at 18:52
user35443
201213
asked Jan 8 at 12:33
jessg12345jessg12345
64
edited Jan 8 at 18:52
user35443
201213
asked Jan 8 at 12:33
jessg12345jessg12345
64
edited Jan 8 at 18:52
user35443
201213
edited Jan 8 at 18:52
user35443
201213
edited Jan 8 at 18:52
user35443
201213
201213
asked Jan 8 at 12:33
jessg12345jessg12345
64
asked Jan 8 at 12:33
jessg12345jessg12345
64
asked Jan 8 at 12:33
jessg12345jessg12345
64
64
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
The region $U$ is in fact an infinite horizontal strip in the upper complex plane.
The conformal transformation $z mapsto exp(pi z)$ will put you to the upper complex plane, and then composition with the map $z mapsto frac{z - i}{z+i}$ gives you the unit disc.
So, we're talking about a conformal transformation $f: U rightarrow mathbb{C}$
$$
f(z) = frac{exp(pi z) - i}{exp(pi z)+i}
$$
Now the Transfer lemma says that a composition of any holomorphic function $g: G rightarrow hat{G}$ with a harmonic function $u: hat{G} rightarrow hat{G}$ will give a function $(u circ g): G rightarrow hat{G}$ that is harmonic. Thus composing $f$ (as conformal maps are holomorphic) with a solution $u$ (to the Dirichlet problem) in the unit disc will yield the result.
$$
u(w) = frac{1}{2 pi}
int_{0}^{pi}{frac{1 - left|frac{exp(pi z) - i}{exp(pi z)+i}right|^2}{
left|exp(itheta) - frac{exp(pi z) - i}{exp(pi z)+i}right|^2
}}text{d}theta
$$
for any $w$ in the $U$ as desired. Notice the change in the interval of integration. This is due to our boundary condition vanishing on the lower semi-circle (counter-clockwise) and being $1$ on the upper semi-circle (clockwise). To figure this out, one needs to keep track of the boundaries as the conformal maps stack up.
In general, the Riemann mapping theorem ensures that any simply connected region can be mapped to the unit disc with a map that is bijective, conformal and has a conformal inverse. So generally, unless the given boundary condition tends to misbehave on the boundary (being at least piece-wise continuous on the boundary still ensures proper behavior IIRC), for any such problem you might be given, the only thing you need to do is to be able to find a conformal mapping from the region given and not lose track of the boundary in the process.
answered Jan 8 at 16:30
user35443user35443
201213
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
The region $U$ is in fact an infinite horizontal strip in the upper complex plane.
The conformal transformation $z mapsto exp(pi z)$ will put you to the upper complex plane, and then composition with the map $z mapsto frac{z - i}{z+i}$ gives you the unit disc.
So, we're talking about a conformal transformation $f: U rightarrow mathbb{C}$
$$
f(z) = frac{exp(pi z) - i}{exp(pi z)+i}
$$
Now the Transfer lemma says that a composition of any holomorphic function $g: G rightarrow hat{G}$ with a harmonic function $u: hat{G} rightarrow hat{G}$ will give a function $(u circ g): G rightarrow hat{G}$ that is harmonic. Thus composing $f$ (as conformal maps are holomorphic) with a solution $u$ (to the Dirichlet problem) in the unit disc will yield the result.
$$
u(w) = frac{1}{2 pi}
int_{0}^{pi}{frac{1 - left|frac{exp(pi z) - i}{exp(pi z)+i}right|^2}{
left|exp(itheta) - frac{exp(pi z) - i}{exp(pi z)+i}right|^2
}}text{d}theta
$$
for any $w$ in the $U$ as desired. Notice the change in the interval of integration. This is due to our boundary condition vanishing on the lower semi-circle (counter-clockwise) and being $1$ on the upper semi-circle (clockwise). To figure this out, one needs to keep track of the boundaries as the conformal maps stack up.
In general, the Riemann mapping theorem ensures that any simply connected region can be mapped to the unit disc with a map that is bijective, conformal and has a conformal inverse. So generally, unless the given boundary condition tends to misbehave on the boundary (being at least piece-wise continuous on the boundary still ensures proper behavior IIRC), for any such problem you might be given, the only thing you need to do is to be able to find a conformal mapping from the region given and not lose track of the boundary in the process.
answered Jan 8 at 16:30
user35443user35443
201213
$endgroup$
add a comment |
$begingroup$
The region $U$ is in fact an infinite horizontal strip in the upper complex plane.
The conformal transformation $z mapsto exp(pi z)$ will put you to the upper complex plane, and then composition with the map $z mapsto frac{z - i}{z+i}$ gives you the unit disc.
So, we're talking about a conformal transformation $f: U rightarrow mathbb{C}$
$$
f(z) = frac{exp(pi z) - i}{exp(pi z)+i}
$$
Now the Transfer lemma says that a composition of any holomorphic function $g: G rightarrow hat{G}$ with a harmonic function $u: hat{G} rightarrow hat{G}$ will give a function $(u circ g): G rightarrow hat{G}$ that is harmonic. Thus composing $f$ (as conformal maps are holomorphic) with a solution $u$ (to the Dirichlet problem) in the unit disc will yield the result.
$$
u(w) = frac{1}{2 pi}
int_{0}^{pi}{frac{1 - left|frac{exp(pi z) - i}{exp(pi z)+i}right|^2}{
left|exp(itheta) - frac{exp(pi z) - i}{exp(pi z)+i}right|^2
}}text{d}theta
$$
for any $w$ in the $U$ as desired. Notice the change in the interval of integration. This is due to our boundary condition vanishing on the lower semi-circle (counter-clockwise) and being $1$ on the upper semi-circle (clockwise). To figure this out, one needs to keep track of the boundaries as the conformal maps stack up.
In general, the Riemann mapping theorem ensures that any simply connected region can be mapped to the unit disc with a map that is bijective, conformal and has a conformal inverse. So generally, unless the given boundary condition tends to misbehave on the boundary (being at least piece-wise continuous on the boundary still ensures proper behavior IIRC), for any such problem you might be given, the only thing you need to do is to be able to find a conformal mapping from the region given and not lose track of the boundary in the process.
answered Jan 8 at 16:30
user35443user35443
201213
$endgroup$
add a comment |
$begingroup$
The region $U$ is in fact an infinite horizontal strip in the upper complex plane.
The conformal transformation $z mapsto exp(pi z)$ will put you to the upper complex plane, and then composition with the map $z mapsto frac{z - i}{z+i}$ gives you the unit disc.
So, we're talking about a conformal transformation $f: U rightarrow mathbb{C}$
$$
f(z) = frac{exp(pi z) - i}{exp(pi z)+i}
$$
Now the Transfer lemma says that a composition of any holomorphic function $g: G rightarrow hat{G}$ with a harmonic function $u: hat{G} rightarrow hat{G}$ will give a function $(u circ g): G rightarrow hat{G}$ that is harmonic. Thus composing $f$ (as conformal maps are holomorphic) with a solution $u$ (to the Dirichlet problem) in the unit disc will yield the result.
$$
u(w) = frac{1}{2 pi}
int_{0}^{pi}{frac{1 - left|frac{exp(pi z) - i}{exp(pi z)+i}right|^2}{
left|exp(itheta) - frac{exp(pi z) - i}{exp(pi z)+i}right|^2
}}text{d}theta
$$
for any $w$ in the $U$ as desired. Notice the change in the interval of integration. This is due to our boundary condition vanishing on the lower semi-circle (counter-clockwise) and being $1$ on the upper semi-circle (clockwise). To figure this out, one needs to keep track of the boundaries as the conformal maps stack up.
In general, the Riemann mapping theorem ensures that any simply connected region can be mapped to the unit disc with a map that is bijective, conformal and has a conformal inverse. So generally, unless the given boundary condition tends to misbehave on the boundary (being at least piece-wise continuous on the boundary still ensures proper behavior IIRC), for any such problem you might be given, the only thing you need to do is to be able to find a conformal mapping from the region given and not lose track of the boundary in the process.
answered Jan 8 at 16:30
user35443user35443
201213
$endgroup$
The region $U$ is in fact an infinite horizontal strip in the upper complex plane.
The conformal transformation $z mapsto exp(pi z)$ will put you to the upper complex plane, and then composition with the map $z mapsto frac{z - i}{z+i}$ gives you the unit disc.
So, we're talking about a conformal transformation $f: U rightarrow mathbb{C}$
$$
f(z) = frac{exp(pi z) - i}{exp(pi z)+i}
$$
Now the Transfer lemma says that a composition of any holomorphic function $g: G rightarrow hat{G}$ with a harmonic function $u: hat{G} rightarrow hat{G}$ will give a function $(u circ g): G rightarrow hat{G}$ that is harmonic. Thus composing $f$ (as conformal maps are holomorphic) with a solution $u$ (to the Dirichlet problem) in the unit disc will yield the result.
$$
u(w) = frac{1}{2 pi}
int_{0}^{pi}{frac{1 - left|frac{exp(pi z) - i}{exp(pi z)+i}right|^2}{
left|exp(itheta) - frac{exp(pi z) - i}{exp(pi z)+i}right|^2
}}text{d}theta
$$
for any $w$ in the $U$ as desired. Notice the change in the interval of integration. This is due to our boundary condition vanishing on the lower semi-circle (counter-clockwise) and being $1$ on the upper semi-circle (clockwise). To figure this out, one needs to keep track of the boundaries as the conformal maps stack up.
In general, the Riemann mapping theorem ensures that any simply connected region can be mapped to the unit disc with a map that is bijective, conformal and has a conformal inverse. So generally, unless the given boundary condition tends to misbehave on the boundary (being at least piece-wise continuous on the boundary still ensures proper behavior IIRC), for any such problem you might be given, the only thing you need to do is to be able to find a conformal mapping from the region given and not lose track of the boundary in the process.
answered Jan 8 at 16:30
user35443user35443
201213
edited Jan 8 at 16:40
answered Jan 8 at 16:30
user35443user35443
201213
answered Jan 8 at 16:30
user35443user35443
201213
answered Jan 8 at 16:30
user35443user35443
201213
201213
add a comment |
add a comment |
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