SpECTRE Documentation Coverage Report
Current view: top level - Domain/CoordinateMaps - FlatOffsetWedge.hpp Coverage Total Hit
Commit: 0d1d717af84a81fb1284a20c4013708fe85cc2f0 Lines: 18.2 % 22 4
Test Date: 2026-10-22 17:29:42
Legend: Lines:     hit not hit

            Line data    Source code
       1            1 : // Distributed under the MIT License.
       2              : // See LICENSE.txt for details.
       3              : 
       4              : /// \file
       5              : /// Defines the class FlattOffsetWedge.
       6              : 
       7              : #pragma once
       8              : 
       9              : #include <array>
      10              : #include <cstddef>
      11              : #include <limits>
      12              : #include <optional>
      13              : 
      14              : #include "DataStructures/Tensor/TypeAliases.hpp"
      15              : 
      16              : /// \cond
      17              : namespace PUP {
      18              : class er;
      19              : }  // namespace PUP
      20              : /// \endcond
      21              : 
      22              : namespace domain::CoordinateMaps {
      23              : 
      24              : /*!
      25              :  * \ingroup CoordinateMapsGroup
      26              :  *
      27              :  * \brief Map from a cube to a volume that connects three planes and
      28              :  * a portion of one spherical surface.
      29              :  *
      30              :  * \image html FlatOffsetWedge.svg "Two slices through a FlatOffsetWedge."
      31              :  *
      32              :  * \details A cube is mapped to the volume shown in the figure.
      33              :  *
      34              :  * A $y=0$ slice through the volume is shown in the left
      35              :  * panel of the figure, and a $x=$const slice through the volume is shown
      36              :  * in the right panel of the figure.
      37              :  *
      38              :  * The upper-$z$ face of the volume is a portion of a spherical surface
      39              :  * of radius $R$ centered about point $C$ in the figure, which is along
      40              :  * the $x$-axis a distance $D$ from the origin (point $O$ in the figure).
      41              :  * The lower-$z$ face of the volume is a two-dimensional rectangle of
      42              :  * constant $z$, located a distance $L$ above the origin.
      43              :  * This rectangle extends from
      44              :  * $0 \leq x \leq D$ in the $x$ direction, and from
      45              :  * $-L \leq y \leq+L$ in the $y$ direction.
      46              :  *
      47              :  * The lower-$x$ face of the volume is a portion of the plane at constant
      48              :  * $x=0$, and the upper-$x$ face of the volume is a portion of the plane
      49              :  * at constant $x=D$. See the $x$ extents of the left panel of the
      50              :  * figure.
      51              :  *
      52              :  * Every constant-$x$ cross section of the volume looks like the right panel
      53              :  * of the figure: it is a two-dimensional wedge with 45 degree
      54              :  * opening angle and a base of
      55              :  * width $2L$.  The outer radius of this wedge varies with $x$ (this radius
      56              :  * is $R$ at $x=D$ and $\sqrt{R^2-D^2}$ at $x=0$).
      57              :  *
      58              :  * ### Restrictions
      59              :  *
      60              :  * We require $D^2 + L^2 < R^2$ or else the lower-$x$ face of the mapped
      61              :  * volume lies outside of the sphere of radius $R$.  This implies $L<R$
      62              :  * and $D<R$.
      63              :  * However, the condition that the upper-$z$ and lower-$z$ surface don't touch
      64              :  * each other at the $\xi=-1$, $\eta=\pm 1$ corners is more stringent:
      65              :  * $D^2 + 2L^2 < R^2$.  This implies that $L<R/\sqrt{2}$.
      66              :  *
      67              :  * ## Equations for the map
      68              :  * Given our cube coordinates $\xi,\eta,\zeta$, each taking on values from
      69              :  * $-1$ to $+1$, we can derive the formulas for the map.
      70              :  *
      71              :  * Define
      72              :  * \begin{equation}
      73              :  *    q \equiv \frac{D}{2R}.
      74              :  * \end{equation}
      75              :  *
      76              :  * Notice that $q<1/2$, because of the restriction $D<R$.
      77              :  *
      78              :  * First, note that $x$ is a function of $\xi$ only, for all points in
      79              :  * the cube:
      80              :  * \begin{align}
      81              :  *  x(\xi,\eta,\zeta) &= q R (\xi+1).
      82              :  * \end{align}
      83              :  *
      84              :  * ### Surface map for bottom and top surfaces
      85              :  * The $y$ and $z$ coordinates for the mapped $\zeta=-1$ surface are
      86              :  * \begin{align}
      87              :  *   \begin{bmatrix}
      88              :  *      y(\xi,\eta,-1)\\
      89              :  *      z(\xi,\eta,-1)
      90              :  *   \end{bmatrix} = L
      91              :  *   \begin{bmatrix}
      92              :  *      \eta\\
      93              :  *      1
      94              :  *   \end{bmatrix}.
      95              :  * \end{align}
      96              :  *
      97              :  * The $y$ and $z$ coordinates of the top surface,
      98              :  * $\zeta=+1$, of the volume are given by
      99              :  * \begin{align}
     100              :  *   \begin{bmatrix}
     101              :  *      y(\xi,\eta,+1)\\
     102              :  *      z(\xi,\eta,+1)
     103              :  *   \end{bmatrix} =
     104              :  *     \frac{R}{\sqrt{1+\eta^2}}
     105              :  *              \sqrt{1-q^2(\xi-1)^2}
     106              :  *   \begin{bmatrix}
     107              :  *      \eta \\
     108              :  *      1
     109              :  *   \end{bmatrix}.
     110              :  * \end{align}
     111              :  *
     112              :  * ### Full volume map
     113              :  * Adding the $\zeta$ dependence by linear interpolation gives us
     114              :  * the full volume map for the $y$ and $z$ coordinates:
     115              :  *
     116              :  * \begin{align}
     117              :  *   \begin{bmatrix}
     118              :  *      y(\xi,\eta,\zeta)\\
     119              :  *      z(\xi,\eta,\zeta)
     120              :  *   \end{bmatrix} =
     121              :  *      \left[L \frac{1-\zeta}{2}
     122              :  *        + \frac{1+\zeta}{2}
     123              :  *          \frac{R}{\sqrt{1+\eta^2}}
     124              :  *           \sqrt{1-q^2(\xi-1)^2}\right]
     125              :  *   \begin{bmatrix}
     126              :  *      \eta \\
     127              :  *      1
     128              :  *   \end{bmatrix}.
     129              :  * \end{align}
     130              :  *
     131              :  * ## Inverse
     132              :  *
     133              :  * The map can be inverted analytically:
     134              :  *
     135              :  * \begin{align}
     136              :  *    \xi   &= \frac{x}{qR}-1\\
     137              :  *    \eta  &= \frac{y}{z}\\
     138              :  *    \zeta &= \frac{2z - L - P}{P-L},
     139              :  *              \label{eq:zetaFromxyz}
     140              :  * \end{align}
     141              :  * where
     142              :  * \begin{align}
     143              :  *   P &\equiv R\frac{\sqrt{1-(x-D)^2/R^2}}{\sqrt{1+y^2/z^2}}\\
     144              :  *     &= R\frac{\sqrt{1-q^2(\xi-1)^2}}{\sqrt{1+\eta^2}}.
     145              :  *              \label{eq:Pdefinition}
     146              :  * \end{align}
     147              :  *
     148              :  * It is easy to determine whether a point $(x,y,z)$ lies within the volume:
     149              :  * First compute $\xi$ and $\eta$ and check that they are both in $[-1,1]$.
     150              :  * If so, then the arguments of the square roots in
     151              :  * Eq. ($\ref{eq:Pdefinition}$) are guaranteed positive, and the denominator
     152              :  * of Eq. ($\ref{eq:zetaFromxyz}$) is guaranteed positive
     153              :  * by our condition $R^2>2L^2+D^2$,
     154              :  * so it is straightforward to
     155              :  * compute $\zeta$ and then check if it is in $[-1,1]$.
     156              :  *
     157              :  * ## Jacobian
     158              :  *
     159              :  * Straightforward differentiation gives
     160              :  *
     161              :  * \begin{align}
     162              :  *    \frac{\partial x}{\partial \xi}   &= qR\\
     163              :  *    \frac{\partial x}{\partial \eta}   &= 0\\
     164              :  *    \frac{\partial x}{\partial \zeta}   &= 0,
     165              :  * \end{align}
     166              :  *
     167              :  * \begin{align}
     168              :  *   \partial_\zeta \begin{bmatrix}
     169              :  *      y\\
     170              :  *      z
     171              :  *   \end{bmatrix} &= \frac{1}{2}
     172              :  *    \left[\frac{R}{\sqrt{1+\eta^2}}\sqrt{1-q^2(\xi-1)^2}
     173              :  *    -L\right]
     174              :  *   \begin{bmatrix}
     175              :  *      \eta \\
     176              :  *      1
     177              :  *   \end{bmatrix}, \\
     178              :  *   \partial_\xi \begin{bmatrix}
     179              :  *      y\\
     180              :  *      z
     181              :  *   \end{bmatrix} &=
     182              :  *    \left[\frac{Rq^2 (1+\zeta)(1-\xi)}{2\sqrt{1+\eta^2}}
     183              :  *    \left(1-q^2(\xi-1)^2\right)^{-1/2}\right]
     184              :  *   \begin{bmatrix}
     185              :  *      \eta \\
     186              :  *      1
     187              :  *   \end{bmatrix}, \\
     188              :  *   \partial_\eta \begin{bmatrix}
     189              :  *      y\\
     190              :  *      z
     191              :  *   \end{bmatrix} &=
     192              :  *       -\left[\frac{R\eta(1+\zeta)}{2(1+\eta^2)^{3/2}}
     193              :  *        \sqrt{1-q^2(\xi-1)^2}\right]
     194              :  *   \begin{bmatrix}
     195              :  *      \eta \\
     196              :  *      1
     197              :  *   \end{bmatrix}+
     198              :  *   \begin{bmatrix}
     199              :  *      z \\
     200              :  *      0
     201              :  *   \end{bmatrix}.
     202              :  * \end{align}
     203              :  *
     204              :  * ## Inverse Jacobian
     205              :  *
     206              :  * Let
     207              :  * \begin{align}
     208              :  *   \Xi &\equiv P\frac{1+\zeta}{P-L}.
     209              :  * \end{align}
     210              :  * Then
     211              :  * \begin{align}
     212              :  *    \frac{\partial \zeta}{\partial x}  &=
     213              :  *        \Xi q\frac{\xi-1}{R\sqrt{1-q^2(\xi-1)^2}},\\
     214              :  *    \frac{\partial \zeta}{\partial y}  &=
     215              :  *        \Xi \frac{\eta}{z\sqrt{1+\eta^2}},\\
     216              :  *    \frac{\partial \zeta}{\partial z}  &=
     217              :  *        -\Xi \eta\frac{\eta}{z\sqrt{1+\eta^2}}
     218              :  *     +\frac{2}{P-L},\\
     219              :  *    \frac{\partial \xi}{\partial x}   &= \frac{1}{qR}\\
     220              :  *    \frac{\partial \xi}{\partial y}   &= 0\\
     221              :  *    \frac{\partial \xi}{\partial z}   &= 0\\
     222              :  *    \frac{\partial \eta}{\partial x}  &= 0\\
     223              :  *    \frac{\partial \eta}{\partial y}  &= \frac{1}{z}\\
     224              :  *    \frac{\partial \eta}{\partial z}  &= -\frac{\eta}{z}.
     225              :  * \end{align}
     226              :  *
     227              :  */
     228            1 : class FlatOffsetWedge {
     229              :  public:
     230            0 :   static constexpr size_t dim = 3;
     231              :   /*!
     232              :    * \brief Constructs a FlatOffsetWedge.
     233              :    *
     234              :    * \param lower_face_y_half_width The half-width $L$ of the lower face in
     235              :    * the $y$ direction.
     236              :    * \param lower_face_x_width The width $D$ of the lower face in
     237              :    * the $x$ direction.
     238              :    * \param outer_radius The outer radius $R$.
     239              :    */
     240            1 :   FlatOffsetWedge(double lower_face_y_half_width, double lower_face_x_width,
     241              :                   double outer_radius);
     242              : 
     243            0 :   FlatOffsetWedge() = default;
     244            0 :   ~FlatOffsetWedge() = default;
     245            0 :   FlatOffsetWedge(FlatOffsetWedge&&) = default;
     246            0 :   FlatOffsetWedge(const FlatOffsetWedge&) = default;
     247            0 :   FlatOffsetWedge& operator=(const FlatOffsetWedge&) = default;
     248            0 :   FlatOffsetWedge& operator=(FlatOffsetWedge&&) = default;
     249              : 
     250              :   template <typename T>
     251            0 :   std::array<T, 3> operator()(const std::array<T, 3>& source_coords) const;
     252              : 
     253              :   /// The inverse function is only callable with doubles because the inverse
     254              :   /// might fail if called for a point out of range, and it is unclear
     255              :   /// what should happen if the inverse were to succeed for some points in a
     256              :   /// DataVector but fail for other points.
     257            1 :   std::optional<std::array<double, 3>> inverse(
     258              :       const std::array<double, 3>& target_coords) const;
     259              : 
     260              :   template <typename T>
     261            0 :   tnsr::Ij<T, 3, Frame::NoFrame> jacobian(
     262              :       const std::array<T, 3>& source_coords) const;
     263              : 
     264              :   template <typename T>
     265            0 :   tnsr::Ij<T, 3, Frame::NoFrame> inv_jacobian(
     266              :       const std::array<T, 3>& source_coords) const;
     267              : 
     268              :   // clang-tidy: google runtime references
     269            0 :   void pup(PUP::er& p);  // NOLINT
     270              : 
     271            0 :   static bool is_identity() { return false; }
     272              : 
     273            0 :   static constexpr bool supports_hessian{false};
     274              : 
     275              :  private:
     276            0 :   friend bool operator==(const FlatOffsetWedge& lhs,
     277              :                          const FlatOffsetWedge& rhs);
     278            0 :   double lower_face_y_half_width_{std::numeric_limits<double>::signaling_NaN()};
     279            0 :   double lower_face_x_width_{std::numeric_limits<double>::signaling_NaN()};
     280            0 :   double outer_radius_{std::numeric_limits<double>::signaling_NaN()};
     281              : };
     282              : 
     283            0 : bool operator!=(const FlatOffsetWedge& lhs, const FlatOffsetWedge& rhs);
     284              : }  // namespace domain::CoordinateMaps
        

Generated by: LCOV version 2.0-1