Benchmark solution
=====================


Calculation method used for the reference solution
--------------------------------------------------------


Modeling A
~~~~~~~~~~~~~~~

The test focuses on the value of the error indicator at the output of RAFF_XFEM. Note :math:`I(M)` the value of this indicator at any point :math:`M`.

At point :math:`{P}_{1}` at the bottom left of the structure, the indicator is the opposite of the distance to the nearest notch point, which is :math:`I({P}_{1})\mathrm{=}\mathrm{-}({\mathit{CP}}_{1}\mathrm{-}{R}_{e})`.

At point :math:`{P}_{2}` at the bottom right of the structure, the indicator is the opposite of the distance to the nearest notch point, which is :math:`I({P}_{2})\mathrm{=}\mathrm{-}({\mathit{DP}}_{2}\mathrm{-}{R}_{e})`.

At point :math:`{P}_{3}` at the top right of the structure, the indicator is the opposite of the distance to the right end of the rightmost crack, i.e. :math:`I({P}_{3})\mathrm{=}\mathrm{-}A\text{'}{P}_{3}` or :math:`A\text{'}\mathrm{=}A+\frac{{L}_{A}}{2}\overrightarrow{x}`.

At point :math:`{P}_{4}` at the top left of the structure, the indicator is the opposite of the distance to the left end of the leftmost crack, i.e. :math:`I({P}_{4})\mathrm{=}\mathrm{-}B\text{'}{P}_{4}` or :math:`B\text{'}\mathrm{=}B\mathrm{-}\frac{{L}_{B}}{2}\overrightarrow{x}`.


B and C models
~~~~~~~~~~~~~~~~~~~~~~

The test focuses on the value of the diameter of the smallest mesh. If :math:`{h}_{0}` is the initial mesh size, :math:`{h}_{c}` is the target mesh size after refinement, then the minimum number of calls to Lobster to reach :math:`{h}_{c}` is :math:`\text{nb\_raff}\mathrm{=}E(n)+1`, with :math:`n\mathrm{=}\frac{\mathrm{ln}({h}_{0})\mathrm{-}\mathrm{ln}({h}_{c})}{\mathrm{ln}(2)}`. After refinement, the most refined mesh size is :math:`h\mathrm{=}\frac{{h}_{0}}{{2}^{\text{nb\_raff}}}`.


Benchmark results
----------------------


Modeling A
~~~~~~~~~~~~~~~

With the numerical values used in the test, we find:

:math:`\begin{array}{c}I({P}_{1})\mathrm{=}\mathrm{-}(\sqrt{{\mathrm{0,25}}^{2}+{\mathrm{0,2}}^{2}}\mathrm{-}\mathrm{0,05})\mathrm{\approx }\mathrm{-}\mathrm{0,27015621187164246}\\ I({P}_{2})\mathrm{=}\mathrm{-}(\sqrt{{\mathrm{0,25}}^{2}+{\mathrm{0,2}}^{2}}\mathrm{-}\mathrm{0,05})\mathrm{\approx }\mathrm{-}\mathrm{0,27015621187164246}\\ I({P}_{3})\mathrm{=}\mathrm{-}\sqrt{{\mathrm{0,25}}^{2}+{\mathrm{0,1}}^{2}}\mathrm{\approx }\mathrm{-}\mathrm{0,26925824035672524}\\ I({P}_{4})\mathrm{=}\mathrm{-}\sqrt{{\mathrm{0,25}}^{2}+{\mathrm{0,2}}^{2}}\mathrm{\approx }\mathrm{-}\mathrm{0,32015621187164245}\end{array}`


B and C models
~~~~~~~~~~~~~~~~~~~~~~

With :math:`{h}_{0}\mathrm{=}\frac{\sqrt{(2)}}{20}` and :math:`{h}_{c}\mathrm{=}\frac{{h}_{0}}{10}`, we get :math:`h\mathrm{=}\mathrm{0,0044194}`.