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


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

In this problem, the boundary conditions are adiabatic, the initial temperature is constant equal to :math:`{T}_{0}` and the load is reduced to the heat source as a function of the temperature :math:`r(T)\mathrm{=}{r}_{0}\mathrm{-}{r}_{1}T` where :math:`{r}_{1}` is positive for reasons of thermal stability. These conditions ensure a homogeneous solution in space. The heat equation is reduced to:

 :math:`\rho {C}_{p}\dot{T}\mathrm{=}{r}_{0}\mathrm{-}{r}_{1}T`; :math:`T(0)\mathrm{=}{T}_{0}` [:ref:`éq1 <éq1>`]

By normalization, we can reduce ourselves without loss of generality to the following equation:

 :math:`\dot{u}\mathrm{=}1\mathrm{-}\omega u`; :math:`u(0)\mathrm{=}0` [:ref:`éq2 <éq2>`]

The solution to this first-order differential equation is then:

 :math:`u(t)\mathrm{=}\frac{1}{\omega }(1\mathrm{-}{e}^{\mathrm{-}\omega t})`

Rather than going back from :math:`u` solution of [:ref:`éq2 <éq2>`] to :math:`T` solution of [:ref:`éq1 <éq1>`], we prefer to adopt the following set of parameters, without paying attention to the units, which leads to :math:`T\mathrm{=}u`: :math:`{T}_{0}\mathrm{=}0`, :math:`{r}_{0}\mathrm{=}\rho C` and :math:`{r}_{1}\mathrm{=}\omega {r}_{0}`.


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

The test case is conducted with :math:`\omega \mathrm{=}2` and the temperature at :math:`t=1` is examined at any node in the element. The data is as follows:


.. csv-table::

    "Thermal conductivity", "LAMBDA ", "0."
    "Volume heat capacity", "RHO_CP ", "2."
    "Initial temperature", ":math:`{T}_{0}` ", "0."
    "Heat source", ":math:`{r}_{0}`
    :math:`{r}_{1}` ", "2.
    4."



.. csv-table::

    "Size tested", ":math:`T` (:math:`t\mathrm{=}1`)"
    "Reference value", ":math:`0.432332`"