Mapping a curve-sided quadrilateral to a rectangle
$begingroup$
I am currently investigating different ways of solving the Laplace equation
$$frac{partial^2 F}{partial x^2} + frac{partial^2 F}{partial z^2} = 0 $$
numerically on the domain $Omega$ shown as the shaded region in the figure below. As the figure shows, $Omega$ is bounded by the vertical lines $x = 0$, $x = L$, the horizontal line $z = -h$ and the curve $(x, eta(x))$ where $eta(x)$ is a smooth, $L$-periodic function.
One strategy for solving the Laplace equation is to make a coordinate transformation $(x,z) mapsto (r(x,z), s(x,z))$ and then solve the equation that the function $G(r(x,z), s(x,z)) = F(x,z)$ satisfies on the transformed domain. For this strategy to be practical, the coordinate transformation should map $Omega$ to a rectangle.
Now, my questions (which is probably more of a request) is the following: Which coordinate tranformations from $Omega$ to a rectangle exist?
I am aware of the simple vertical stretching $(x,z) mapsto (x, s(x,z))$ where
$$ s(x,z) = frac{2z + h - eta(x)}{h + eta(x)}$$
and in principal it works fine. In the mapped coordinates the function $G(x,s(x,z)) = F(x,z)$ solves the equation
$$ frac{partial^2 G}{partial x^2} + bigg( bigg[frac{partial s}{partial x} bigg]^2 + bigg[ frac{partial s}{partial z} bigg]^2 bigg) frac{partial^2 G}{partial s^2} + 2 frac{partial s}{partial x} frac{partial^2 G}{partial x partial s} + frac{partial^2 s}{partial x^2} frac{partial G}{partial s} = 0, qquad 0 leq x leq L, , , , -1 leq s leq 1$$
and while this certainly can be solved numerically, it would be nice to have a coordinate transformation which does not give rise to such a complicated equation. The ideal transformation would of course be a conformal one, since the Laplace equation is invariant under such a change of coordinates.
differential-geometry pde harmonic-functions conformal-geometry
asked Jan 3 at 12:53
Mathias KlahnMathias Klahn
362
$endgroup$
add a comment |
$begingroup$
I am currently investigating different ways of solving the Laplace equation
$$frac{partial^2 F}{partial x^2} + frac{partial^2 F}{partial z^2} = 0 $$
numerically on the domain $Omega$ shown as the shaded region in the figure below. As the figure shows, $Omega$ is bounded by the vertical lines $x = 0$, $x = L$, the horizontal line $z = -h$ and the curve $(x, eta(x))$ where $eta(x)$ is a smooth, $L$-periodic function.
One strategy for solving the Laplace equation is to make a coordinate transformation $(x,z) mapsto (r(x,z), s(x,z))$ and then solve the equation that the function $G(r(x,z), s(x,z)) = F(x,z)$ satisfies on the transformed domain. For this strategy to be practical, the coordinate transformation should map $Omega$ to a rectangle.
Now, my questions (which is probably more of a request) is the following: Which coordinate tranformations from $Omega$ to a rectangle exist?
I am aware of the simple vertical stretching $(x,z) mapsto (x, s(x,z))$ where
$$ s(x,z) = frac{2z + h - eta(x)}{h + eta(x)}$$
and in principal it works fine. In the mapped coordinates the function $G(x,s(x,z)) = F(x,z)$ solves the equation
$$ frac{partial^2 G}{partial x^2} + bigg( bigg[frac{partial s}{partial x} bigg]^2 + bigg[ frac{partial s}{partial z} bigg]^2 bigg) frac{partial^2 G}{partial s^2} + 2 frac{partial s}{partial x} frac{partial^2 G}{partial x partial s} + frac{partial^2 s}{partial x^2} frac{partial G}{partial s} = 0, qquad 0 leq x leq L, , , , -1 leq s leq 1$$
and while this certainly can be solved numerically, it would be nice to have a coordinate transformation which does not give rise to such a complicated equation. The ideal transformation would of course be a conformal one, since the Laplace equation is invariant under such a change of coordinates.
differential-geometry pde harmonic-functions conformal-geometry
asked Jan 3 at 12:53
Mathias KlahnMathias Klahn
362
$endgroup$
$begingroup$
The boundaries are well defined. But no boundary condition is given. So they are an infinity of solutions. For example $F(x,y)=f(x+iy)+g(x-iy)$ with arbitrary fonctions $f$ and $g$. All these solutions are valid on your boundaries and are valid on any other boundaries until boundary conditions be specified. Thus, with the present wording of the question the answer is trivially $F(x,y)=f(x+iy)+g(x-iy)$.
$endgroup$
– JJacquelin
Jan 7 at 8:38
$begingroup$
The boundaries that I will eventually impose on the solution are the following: 1) $F$ is periodic in $x$ with period $L$, 2) $F(x, eta(x)) = F_s(x)$ where $F_s(x)$ is some smooth periodic function and 3) $frac{partial F}{partial y} (x, -h) = 0$. At the present time I am however mostly interested in how $Omega$ can be mapped to a rectangle, and this should not depend on the boundary conditions.
$endgroup$
– Mathias Klahn
Jan 7 at 12:05
add a comment |
$begingroup$
I am currently investigating different ways of solving the Laplace equation
$$frac{partial^2 F}{partial x^2} + frac{partial^2 F}{partial z^2} = 0 $$
numerically on the domain $Omega$ shown as the shaded region in the figure below. As the figure shows, $Omega$ is bounded by the vertical lines $x = 0$, $x = L$, the horizontal line $z = -h$ and the curve $(x, eta(x))$ where $eta(x)$ is a smooth, $L$-periodic function.
One strategy for solving the Laplace equation is to make a coordinate transformation $(x,z) mapsto (r(x,z), s(x,z))$ and then solve the equation that the function $G(r(x,z), s(x,z)) = F(x,z)$ satisfies on the transformed domain. For this strategy to be practical, the coordinate transformation should map $Omega$ to a rectangle.
Now, my questions (which is probably more of a request) is the following: Which coordinate tranformations from $Omega$ to a rectangle exist?
I am aware of the simple vertical stretching $(x,z) mapsto (x, s(x,z))$ where
$$ s(x,z) = frac{2z + h - eta(x)}{h + eta(x)}$$
and in principal it works fine. In the mapped coordinates the function $G(x,s(x,z)) = F(x,z)$ solves the equation
$$ frac{partial^2 G}{partial x^2} + bigg( bigg[frac{partial s}{partial x} bigg]^2 + bigg[ frac{partial s}{partial z} bigg]^2 bigg) frac{partial^2 G}{partial s^2} + 2 frac{partial s}{partial x} frac{partial^2 G}{partial x partial s} + frac{partial^2 s}{partial x^2} frac{partial G}{partial s} = 0, qquad 0 leq x leq L, , , , -1 leq s leq 1$$
and while this certainly can be solved numerically, it would be nice to have a coordinate transformation which does not give rise to such a complicated equation. The ideal transformation would of course be a conformal one, since the Laplace equation is invariant under such a change of coordinates.
differential-geometry pde harmonic-functions conformal-geometry
asked Jan 3 at 12:53
Mathias KlahnMathias Klahn
362
$endgroup$
I am currently investigating different ways of solving the Laplace equation
$$frac{partial^2 F}{partial x^2} + frac{partial^2 F}{partial z^2} = 0 $$
numerically on the domain $Omega$ shown as the shaded region in the figure below. As the figure shows, $Omega$ is bounded by the vertical lines $x = 0$, $x = L$, the horizontal line $z = -h$ and the curve $(x, eta(x))$ where $eta(x)$ is a smooth, $L$-periodic function.
One strategy for solving the Laplace equation is to make a coordinate transformation $(x,z) mapsto (r(x,z), s(x,z))$ and then solve the equation that the function $G(r(x,z), s(x,z)) = F(x,z)$ satisfies on the transformed domain. For this strategy to be practical, the coordinate transformation should map $Omega$ to a rectangle.
Now, my questions (which is probably more of a request) is the following: Which coordinate tranformations from $Omega$ to a rectangle exist?
I am aware of the simple vertical stretching $(x,z) mapsto (x, s(x,z))$ where
$$ s(x,z) = frac{2z + h - eta(x)}{h + eta(x)}$$
and in principal it works fine. In the mapped coordinates the function $G(x,s(x,z)) = F(x,z)$ solves the equation
$$ frac{partial^2 G}{partial x^2} + bigg( bigg[frac{partial s}{partial x} bigg]^2 + bigg[ frac{partial s}{partial z} bigg]^2 bigg) frac{partial^2 G}{partial s^2} + 2 frac{partial s}{partial x} frac{partial^2 G}{partial x partial s} + frac{partial^2 s}{partial x^2} frac{partial G}{partial s} = 0, qquad 0 leq x leq L, , , , -1 leq s leq 1$$
and while this certainly can be solved numerically, it would be nice to have a coordinate transformation which does not give rise to such a complicated equation. The ideal transformation would of course be a conformal one, since the Laplace equation is invariant under such a change of coordinates.
differential-geometry pde harmonic-functions conformal-geometry
differential-geometry pde harmonic-functions conformal-geometry
asked Jan 3 at 12:53
Mathias KlahnMathias Klahn
362
asked Jan 3 at 12:53
Mathias KlahnMathias Klahn
362
asked Jan 3 at 12:53
Mathias KlahnMathias Klahn
362
asked Jan 3 at 12:53
Mathias KlahnMathias Klahn
362
asked Jan 3 at 12:53
Mathias KlahnMathias Klahn
362
362
$begingroup$
The boundaries are well defined. But no boundary condition is given. So they are an infinity of solutions. For example $F(x,y)=f(x+iy)+g(x-iy)$ with arbitrary fonctions $f$ and $g$. All these solutions are valid on your boundaries and are valid on any other boundaries until boundary conditions be specified. Thus, with the present wording of the question the answer is trivially $F(x,y)=f(x+iy)+g(x-iy)$.
$endgroup$
– JJacquelin
Jan 7 at 8:38
$begingroup$
The boundaries that I will eventually impose on the solution are the following: 1) $F$ is periodic in $x$ with period $L$, 2) $F(x, eta(x)) = F_s(x)$ where $F_s(x)$ is some smooth periodic function and 3) $frac{partial F}{partial y} (x, -h) = 0$. At the present time I am however mostly interested in how $Omega$ can be mapped to a rectangle, and this should not depend on the boundary conditions.
$endgroup$
– Mathias Klahn
Jan 7 at 12:05
add a comment |
$begingroup$
The boundaries are well defined. But no boundary condition is given. So they are an infinity of solutions. For example $F(x,y)=f(x+iy)+g(x-iy)$ with arbitrary fonctions $f$ and $g$. All these solutions are valid on your boundaries and are valid on any other boundaries until boundary conditions be specified. Thus, with the present wording of the question the answer is trivially $F(x,y)=f(x+iy)+g(x-iy)$.
$endgroup$
– JJacquelin
Jan 7 at 8:38
$begingroup$
The boundaries that I will eventually impose on the solution are the following: 1) $F$ is periodic in $x$ with period $L$, 2) $F(x, eta(x)) = F_s(x)$ where $F_s(x)$ is some smooth periodic function and 3) $frac{partial F}{partial y} (x, -h) = 0$. At the present time I am however mostly interested in how $Omega$ can be mapped to a rectangle, and this should not depend on the boundary conditions.
$endgroup$
– Mathias Klahn
Jan 7 at 12:05
$begingroup$
The boundaries are well defined. But no boundary condition is given. So they are an infinity of solutions. For example $F(x,y)=f(x+iy)+g(x-iy)$ with arbitrary fonctions $f$ and $g$. All these solutions are valid on your boundaries and are valid on any other boundaries until boundary conditions be specified. Thus, with the present wording of the question the answer is trivially $F(x,y)=f(x+iy)+g(x-iy)$.
$endgroup$
– JJacquelin
Jan 7 at 8:38
$begingroup$
The boundaries are well defined. But no boundary condition is given. So they are an infinity of solutions. For example $F(x,y)=f(x+iy)+g(x-iy)$ with arbitrary fonctions $f$ and $g$. All these solutions are valid on your boundaries and are valid on any other boundaries until boundary conditions be specified. Thus, with the present wording of the question the answer is trivially $F(x,y)=f(x+iy)+g(x-iy)$.
$endgroup$
– JJacquelin
Jan 7 at 8:38
$begingroup$
The boundaries that I will eventually impose on the solution are the following: 1) $F$ is periodic in $x$ with period $L$, 2) $F(x, eta(x)) = F_s(x)$ where $F_s(x)$ is some smooth periodic function and 3) $frac{partial F}{partial y} (x, -h) = 0$. At the present time I am however mostly interested in how $Omega$ can be mapped to a rectangle, and this should not depend on the boundary conditions.
$endgroup$
– Mathias Klahn
Jan 7 at 12:05
$begingroup$
The boundaries that I will eventually impose on the solution are the following: 1) $F$ is periodic in $x$ with period $L$, 2) $F(x, eta(x)) = F_s(x)$ where $F_s(x)$ is some smooth periodic function and 3) $frac{partial F}{partial y} (x, -h) = 0$. At the present time I am however mostly interested in how $Omega$ can be mapped to a rectangle, and this should not depend on the boundary conditions.
$endgroup$
– Mathias Klahn
Jan 7 at 12:05
add a comment |
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$begingroup$
The boundaries are well defined. But no boundary condition is given. So they are an infinity of solutions. For example $F(x,y)=f(x+iy)+g(x-iy)$ with arbitrary fonctions $f$ and $g$. All these solutions are valid on your boundaries and are valid on any other boundaries until boundary conditions be specified. Thus, with the present wording of the question the answer is trivially $F(x,y)=f(x+iy)+g(x-iy)$.
$endgroup$
– JJacquelin
Jan 7 at 8:38
$begingroup$
The boundaries that I will eventually impose on the solution are the following: 1) $F$ is periodic in $x$ with period $L$, 2) $F(x, eta(x)) = F_s(x)$ where $F_s(x)$ is some smooth periodic function and 3) $frac{partial F}{partial y} (x, -h) = 0$. At the present time I am however mostly interested in how $Omega$ can be mapped to a rectangle, and this should not depend on the boundary conditions.
$endgroup$
– Mathias Klahn
Jan 7 at 12:05