Reference solution
=====================


Calculation method
-----------------

It is an analytical solution. In this configuration the problem is one-dimensional and the heat equation is reduced to the following differential equation:

 :math:`\frac{{d}^{2}T}{{\mathit{dy}}^{2}}=0` with conditions :math:`T(y\mathrm{=}0)\mathrm{=}\text{}\stackrel{ˉ}{T}{\text{}}^{\text{inf}}` and :math:`T(y\mathrm{=}\mathit{LY})\mathrm{=}\text{}\stackrel{ˉ}{T}{\text{}}^{\text{sup}}`

This equation has the following solution:

 :math:`T(y)\mathrm{=}(\text{}\stackrel{ˉ}{T}{\text{}}^{\text{sup}}\mathrm{-}\text{}\stackrel{ˉ}{T}{\text{}}^{\text{inf}})\frac{y}{\mathit{LY}}+\text{}\stackrel{ˉ}{T}{\text{}}^{\text{inf}}`


Reference quantities and results
-----------------------------------

The temperature is tested at points :math:`{P}^{\text{+}}`, :math:`{P}^{\text{-}}` and :math:`Q` (see Figure):

 :math:`T(\frac{\mathit{LY}}{2})=\frac{(\text{}\stackrel{̄}{T}{\text{}}^{\text{sup}}+\text{}\stackrel{̄}{T}{\text{}}^{\text{inf}})}{2}`, digital application: :math:`T(\frac{\mathit{LY}}{2})=15°C`

The two components (space constants) of the temperature gradient are also tested:

 :math:`{\partial }_{x}T=0`, digital application: :math:`{\partial }_{x}T=0°{\mathit{C.m}}^{-1}`

 :math:`{\partial }_{y}T=\frac{(\text{}\stackrel{̄}{T}{\text{}}^{\text{sup}}-\text{}\stackrel{̄}{T}{\text{}}^{\text{inf}})}{\mathit{LY}}`, digital application: :math:`{\partial }_{y}T=10°{\mathit{C.m}}^{-1}`


.. csv-table::

    "**Identification**", "**Reference type**", "**Reference value**"
    "Points :math:`{P}^{\text{+}}`, :math:`{P}^{\text{-}}`, and :math:`Q` - :math:`\mathit{TEMP}` ", "'ANALYTIQUE'", ":math:`15°C`"
    "In every way, :math:`\mathit{DTX}` ", "'ANALYTIQUE'", ":math:`0°{\mathit{C.m}}^{-1}`"
    "In every way, :math:`\mathit{DTY}` ", "'ANALYTIQUE'", ":math:`10°{\mathit{C.m}}^{-1}`"