3. Variational formulation of the problem

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

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

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

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

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

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

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

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

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

From this we deduce the variational formulation of the problem:

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