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


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

The bar is subjected to a :math:`r(T)\mathrm{=}{r}_{0}\mathrm{-}{r}_{1}T` heat source, where :math:`{r}_{1}>0` for reasons of thermal stability. Its initial temperature is :math:`{T}_{0}(x)` and the ends of the bar are maintained at zero temperature. The evolution of temperature obeys the heat equation:

:math:`\rho c\dot{T}\mathrm{=}\lambda {\mathrm{\nabla }}^{2}T+r(T)`; :math:`T(x\mathrm{,0})\mathrm{=}{T}_{0}(x)`; :math:`T(\mathrm{-}L,t)\mathrm{=}T(L,t)\mathrm{=}0`

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

:math:`\frac{\mathrm{\partial }u}{\mathrm{\partial }t}\mathrm{=}\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{x}^{2}}+1\mathrm{-}{\omega }^{2}u`; :math:`u(x\mathrm{,0})\mathrm{=}{u}_{0}(x)`; :math:`u(\mathrm{-}\mathrm{1,}t)\mathrm{=}u(\mathrm{1,}t)\mathrm{=}0`

To solve this equation, we first look at the asymptotic solution :math:`{u}_{\mathrm{\infty }}(x)` which verifies:

:math:`0\mathrm{=}\frac{{\mathrm{\partial }}^{2}{u}_{\mathrm{\infty }}}{\mathrm{\partial }{x}^{2}}+1\mathrm{-}{\omega }^{2}{u}_{\mathrm{\infty }}`; :math:`{u}_{\mathrm{\infty }}(\mathrm{-}1)\mathrm{=}{u}_{\mathrm{\infty }}(1)\mathrm{=}0`

The solution of this linear differential equation of the second order is equal to:

:math:`{u}_{\mathrm{\infty }}(x)\mathrm{=}\frac{1}{{\omega }^{2}}(1\mathrm{-}\frac{\mathrm{cosh}\omega x}{\mathrm{cosh}\omega })`

The solution of the transient equation is then obtained by projecting :math:`v\mathrm{=}u\mathrm{-}{u}_{\mathrm{\infty }}` onto the Laplacian eigenfunctions onto :math:`\left\}\mathrm{-}\mathrm{1,}1\right\{`. To simplify the analysis, we adopt an initial condition :math:`{u}_{0}` equal to the first eigenmode, namely:

:math:`{u}_{0}(x)\mathrm{=}{u}_{\mathrm{\infty }}(x)\mathrm{-}\mathrm{cos}\frac{\pi x}{2}`

Only the first mode being activated, we are brought back to the resolution of a first-order time differential equation, to finally obtain the solution:

:math:`u(x,t)\mathrm{=}{u}_{\mathrm{\infty }}(x)\mathrm{-}\mathrm{exp}(\mathrm{-}{\omega }^{2}t\mathrm{-}\frac{{\pi }^{2}}{4}t)\mathrm{cos}\frac{\pi x}{2}`

Finally, as before, we go back from :math:`u` to :math:`T` by adopting a specific set of parameters, without taking into account the units, so that :math:`T\mathrm{=}u`. To do this, we take :math:`\lambda \mathrm{=}{r}_{0}\mathrm{=}\rho c`, :math:`{r}_{1}\mathrm{=}{\omega }^{2}{r}_{0}`,, :math:`L\mathrm{=}1` and :math:`{T}_{0}(x)\mathrm{=}{u}_{0}(x)`.


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

The test case is conducted with :math:`\omega \mathrm{=}\sqrt{2}` and the temperature at :math:`t=1` is examined at a node in the plane of symmetry :math:`(x=0)`. The data is as follows:


.. csv-table::

    "Thermal conductivity", "LAMBDA ", "2."
    "Volume heat capacity", "RHO_CP ", "2."
    "Initial temperature", ":math:`{T}_{0}` "," :math:`{u}_{0}(x)\mathrm{=}\frac{1}{{\omega }^{2}}(1\mathrm{-}\frac{\mathrm{cosh}\omega x}{\mathrm{cosh}\omega })\mathrm{-}\mathrm{cos}\frac{\pi x}{2}` with :math:`\omega \mathrm{=}\sqrt{2}`"
    "Heat source", ":math:`{r}_{0}`
    :math:`{r}_{1}` ", "2.
    4."



.. csv-table::

    "Size tested", ":math:`T` (:math:`x\mathrm{=}\mathrm{0,}t\mathrm{=}1`)"
    "Reference value", ":math:`0.258974`"