Variational formulation of the problem
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Here we will limit ourselves to presenting the problem with only the conditions at the limits of imposed temperature [:external:ref:`R5.02.01 §2.1 <R5.02.01 §2.1>`], imposed normal flow [:external:ref:`R5.02.01 §2.3 <R5.02.01 §2.3>`], exchange [:external:ref:`R5.02.01§2.4 <R5.02.01§2.4>`], non-linear flow [:ref:`§2.1 <§2.1>`] and radiation [:ref:`§2.2 <§2.2>`].

Let :math:`\Omega` be an open of :math:`{R}^{3}`, of border :math:`\Gamma ={\Gamma }_{1}\cup {\Gamma }_{2}\cup {\Gamma }_{3}\cup {\Gamma }_{4}\cup {\Gamma }_{5}`.

We need to solve the equation [:ref:`éq 1.1-4 <éq 1.1-4>`] in :math:`T` over :math:`\Omega \times ]\mathrm{0,}t[` with the boundary conditions:

:math:`\{\begin{array}{ccccc}T={T}^{d}(r,t)& & & \text{sur}& {\Gamma }_{1}\\ \lambda (T)\frac{\partial T}{\partial n}& \text{=}& f(r,t)& \text{sur}& {\Gamma }_{2}\\ \lambda (T)\frac{\partial T}{\partial n}& \text{=}& h(r,t)({T}_{\text{ext}}(r,t)-T)& \text{sur}& {\Gamma }_{3}\\ \lambda (T)\frac{\partial T}{\partial n}& \text{=}& g(r,T)& \text{sur}& {\Gamma }_{4}\\ \lambda (T)\frac{\partial T}{\partial n}& \text{=}& \sigma \epsilon \left[(T+273.15{)}^{4}-({T}_{\infty }+273.15{)}^{4}\right]& \text{sur}& {\Gamma }_{5}\end{array}` eq 3-1

and with, possibly, initial conditions :math:`T(t=0)`. If the latter steps are not specified, the stationary problem is solved beforehand, i.e. equation [:ref:`éq 1.3-1 <éq 1.3-1>`] without the term of temporal evolution.

Let :math:`v` be a sufficiently regular function cancelling out of :math:`{\Gamma }_{1}`, noting:


.. _RefEquation 3-2:

:math:`\begin{array}{}\frac{d}{\text{dt}}(\underset{\Omega }{\int }\beta (T)\mathrm{.}\mathrm{v.d}\Omega )=\underset{\Omega }{\int }\dot{\beta }(T)\mathrm{.}\mathrm{v.d}\Omega \\ \begin{array}{cc}\underset{\Omega }{\int }\lambda (T)\nabla T\text{.}\nabla \mathrm{v.d}\Omega =& -\underset{\Omega }{\int }\text{div}(\lambda (T)\nabla T)\mathrm{.}\mathrm{v.d}\Omega +\underset{\Gamma }{\int }\lambda (T)\frac{\partial T}{\partial n}\mathrm{.}\mathrm{v.d}\Gamma \\ & \end{array}\end{array}` eq 3-2

The weak formulation of the heat equation can then be written as:


.. _RefEquation 3-3:

:math:`\frac{d}{\text{dt}}(\underset{\Omega }{\int }\beta (T)\mathrm{.}\mathrm{v.d}\Omega )+\underset{\Omega }{\int }\lambda (T)\nabla T\text{.}\nabla \mathrm{vd}\Omega -\underset{\Gamma }{\int }\lambda (T)\frac{\partial T}{\partial n}\mathrm{.}\mathrm{v.d}\Gamma =\underset{\Omega }{\int }{r}_{\text{vol}}\mathrm{.}\mathrm{v.d}\Omega` eq 3-3

From this we deduce the variational formulation of the problem:

:math:`\begin{array}{c}\underset{\Omega }{\mathrm{\int }}\frac{d\beta (T)}{\text{dt}}\mathrm{.}\mathit{v.d}\Omega +\underset{\Omega }{\mathrm{\int }}\lambda (T)\mathrm{\nabla }\mathit{T.}\mathrm{\nabla }\mathit{v.d}\Omega +\underset{{\Gamma }_{3}}{\mathrm{\int }}\mathit{hT.v.d}{\Gamma }_{3}\mathrm{=}\\ \underset{\Omega }{\mathrm{\int }}{r}_{\text{vol}}\mathrm{.}\mathit{v.d}\Omega +\underset{{\Gamma }_{2}}{\mathrm{\int }}\mathit{f.v.d}{\Gamma }_{2}+\underset{{\Gamma }_{3}}{\mathrm{\int }}{\mathit{hT}}_{\text{ext}}\text{.}\mathit{vd}{\Gamma }_{3}+\\ \underset{{\Gamma }_{4}}{\mathrm{\int }}\mathit{g.v.d}{\Gamma }_{4}+\underset{{\Gamma }_{5}}{\mathrm{\int }}\sigma \epsilon \mathrm{.}\left[(T+273.15{)}^{4}\mathrm{-}({T}_{\mathrm{\infty }}+273.15{)}^{4}\right]\mathrm{.}\mathit{v.d}{\Gamma }_{5}\end{array}` eq 3-4