Variational formulation of the problem
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Let :math:`\Omega` be an open of :math:`{R}^{3}`, of border :math:`\partial \Omega ={\partial }_{1}\Omega \cup {\partial }_{2}\Omega \cup {\partial }_{3}\Omega \cup {\partial }_{4}\Omega` such that,

For :math:`i\mathrm{\ne }j` and :math:`i,j\mathrm{=}\mathrm{1,}\mathrm{...},4`, we have: :math:`{\partial }_{i}\Omega \cap {\partial }_{j}\Omega \text{=}\varnothing`.

Let :math:`\psi` still be a sufficiently regular function that is cancelled on :math:`{\mathrm{\partial }}_{4}\Omega`: :math:`\psi \mathrm{\in }V\mathrm{=}\left\{y\text{régulière}\text{et}\psi {\mathrm{\mid }}_{{\mathrm{\partial }}_{4}\Omega }\mathrm{=}0\right\}`.

Multiply both sides of the equation [:ref:`éq 2-1 <éq 2-1>`] by :math:`\psi` and then integrate on :math:`\Omega`. An integration by parts then gives:


.. _RefEquation 3-1:

:math:`\begin{array}{ccc}\underset{\Omega }{\int }\mathrm{Qyd}\Omega & \text{=}\underset{\Omega }{\int }V\text{.}\text{grad}u(T)\mathrm{yd}\Omega -& \underset{\Omega }{\int }\text{div}(k(T)\text{grad}T)\mathrm{yd}\Omega \\ & \text{=}\underset{\Omega }{\int }V\text{.}\text{grad}u(T)\mathrm{yd}\Omega +& \underset{\Omega }{\int }k(T)\text{grad}T\text{.}\text{grad}\mathrm{yd}\Omega -\underset{\partial \Omega -{\partial }_{4}\Omega }{\int }(k(T)\frac{\partial T}{\partial n}y)\mathrm{dG}\end{array}` eq 3-1

since :math:`\psi` sucks on :math:`{\partial }_{4}\Omega`.

Hence, taking into account the boundary conditions [:ref:`éq 2-2 <éq 2-2>`], [:ref:`éq 2-3 <éq 2-3>`] and [:ref:`éq 2-4 <éq 2-4>`], the variational formulation of the reference problem which is given by the following equation:

:math:`\forall \psi \in V`


.. _RefEquation 3-2:

:math:`\begin{array}{}\begin{array}{cc}\underset{\Omega }{\int }k(T)\text{grad}T\text{.}\text{grad}\mathrm{yd}\Omega & +\underset{\Omega }{\int }V\text{.}\text{grad}u(T)\mathrm{yd}\Omega +\underset{{\partial }_{1}\Omega }{\int }\mathrm{gTydG}-\underset{{\partial }_{3}\Omega }{\int }\alpha (T)\mathrm{ydG}\end{array}\\ =\underset{\Omega }{\int }\mathrm{Qyd}\Omega +\underset{{\partial }_{1}\Omega }{\int }{\gamma T}_{\text{ext}}\mathrm{ydG}+\underset{{\partial }_{2}\Omega }{\int }{q}_{0}\mathrm{yd}\Omega ,\end{array}` eq 3-2