Analytical solution
===================


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

The monophasic unsteady and one-dimensional problem can be written in a general form such as:

:math:`N\frac{\partial P}{\partial t}-{K}_{i}\Delta P=0`

:math:`P(t=0)={P}_{0}`

:math:`P(t,x=0)=0`

:math:`\frac{\partial P}{\partial x}(t,x,L)=0`

This problem admits an analytical solution obtained by development in Fourier series.

:math:`P={\sum }_{k=0}^{K}\frac{{\mathrm{4P}}_{0}}{(\mathrm{2k}+1)\pi }\mathrm{exp}(\frac{-{K}_{i}}{N}{\omega }_{k}^{2}t)\mathrm{sin}({\omega }_{k}x)` with :math:`{\omega }_{k}=(k+\frac{1}{2})\frac{\pi }{L}`

The number of terms :math:`K` in this series is determined as follows:

Let :math:`{n}_{x}` be the number of :math:`{x}_{i}` points where the solution is evaluated at an instant :math:`t`.

We ask:

:math:`{a}_{k}^{i}=\frac{4}{(\mathrm{2k}+1)\pi }\mathrm{exp}(\frac{-{K}_{i}}{N}{\omega }_{k}^{2}t)\mathrm{sin}({\omega }_{k}{x}_{i})`

So the solution can be written as: :math:`P({x}_{i})={\sum }_{k=0}^{K}{P}_{0}\mathrm{.}{a}_{k}^{i}`

We choose :math:`K` such as: :math:`\frac{1}{{n}_{x}}\sqrt{{\sum }_{i=1}^{\mathrm{nx}}{({a}_{k}^{i})}^{2}}<\epsilon`

In practice we took :math:`\varepsilon ={10}^{-10}`.

The shapes of the analytical solution at times 1, 10, 100, 1000 are shown in the figure below:


.. image:: images/100000000000025800000190C21A5A0074EC48A0.png
    :width: 3.1252in
    :height: 2.0835in


.. _RefImage_100000000000025800000190C21A5A0074EC48A0.png:

The following table shows the number of terms by time:


.. csv-table::

    "Time", "Number of serial terms"
    "1", "194"
    "10", "64"
    "100", "22"
    "1000", "8"


Table 2.1-1: Representation of term number as a function of time


Simplifying hypotheses
---------------------------

It is considered that the medium is completely saturated with gas and a zero liquid pressure is imposed on all nodes in aster. An initial gas pressure :math:`{P}_{g}^{\mathit{ref}}` is imposed and boundary conditions are given which correspond to a variation in this reference pressure, so :math:`\delta {P}_{g}` this pressure variation is given. The gas mass conservation equation will be written as:

:math:`\frac{\partial (\varphi \delta {P}_{g})}{\partial t}+d(\frac{{K}_{i}{k}_{g}}{{\mu }_{g}}({P}_{g}^{r}+\delta {P}_{g})d({P}_{g}^{r}+\delta {P}_{g}))=0`

Assuming :math:`\delta {P}_{g}` small in front of :math:`{P}_{g}^{\mathit{ref}}`, this equation becomes:

:math:`\frac{\partial (\varphi \delta {P}_{g})}{\partial t}+\frac{{K}_{i}{k}_{g}}{{\mu }_{g}}{P}_{g}^{r}\Delta (\delta {P}_{g})=0`

It is therefore :math:`\delta {P}_{g}` that we will identify with the solution of the model analytical equation.

In order to find the coefficients of the model problem, we will take:

:math:`\varphi \mathrm{=}1`

:math:`{k}_{g}\mathrm{=}{\mu }_{g}\mathrm{=}1`

and we will make sure that

:math:`{K}_{\mathit{int}}{P}_{\mathit{ref}}\mathrm{=}{10}^{\mathrm{-}3}`.