.. sectionauthor::   Yasuyuki SHIMIZU <yasu@i-ric.org>

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Appendix III (Diffusion equation in the s-n coordinates)
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.. _sn_diff:

Derivation of the diffusion equation in the s-n coordinates
======================================================================

Let us assume that the :math:`s-n` coordinates are Cartesian curvilinear coordinates. On the :math:`s-n` coordinates, the :math:`s` axis is assumed to be an arbitrary curve, and the :math:`n` axis is assumed to be a straight coordinate axis perpendicular to the `s` axis.

.. _dif_sn:

.. figure:: images/0C/dif_sn.png
    :width: 80%

    : Density flux balance of the infinitesimal elements on the :math:`s-n` coordinates 

Assuming that the density of the diffusion material is :math:`c`, then the unit width flux and the diffusion coefficient in the :math:`s, n` direction are :math:`F_s, F_n` and :math:`D_s, D_n`, respectively. They have following relationships.  

.. math::
    :label: sn_dif1

    F_s=-D_s\cfrac{\partial F_s}{\partial s}, \hspace{2cm} 
    F_n=-D_n\cfrac{\partial F_n}{\partial n}

Considering the density balance of the infinitesimal elements in an infinitesimal time shown in :numref:`dif_sn` and taking into account that the area of the infinitesimal element is :math:`r\, \delta\theta\, \delta n`, then, the following equations are derived. 

.. math::
    :label: sn_dif2

    \cfrac{\partial c}{\partial t} r\, \delta\theta\, \delta n 
    =F_{n(n)}r \delta\theta - F_{n(n+\delta n)} (r+\delta n) \delta\theta
    +F_{s(s)}\delta n -F_{s(s+\delta s)}\delta n

.. math::
    :label: sn_dif21

    \cfrac{\partial c}{\partial t}=\cfrac{F_{n(n)}}{\delta n}-\cfrac{F_{n(n+\delta n)}\, (r+\delta n)}{r\delta n}
    +\cfrac{F_{s(s)}-F_{s(s+\delta s)}}{\delta s}

Where,

.. math::
    :label: sn_dif3

    \cfrac{(r+\delta n)}{r\delta n}=\cfrac{1}{r}+\cfrac{1}{\delta n}

Therefore, the second term on the right side of Equation :eq:`sn_dif21` is... 

.. math::
    :label: sn_dif4

    \cfrac{F_{n(n+\delta n)}\, (r+\delta n)}{r\delta n}=
    \cfrac{F_{n(n+\delta n)}}{r}+\cfrac{F_{n(r+\delta n)}}{\delta n}

Then, Equation :eq:`sn_dif21` is expressed as follows.

.. math::
    :label: sn_dif5

    \cfrac{\partial c}{\partial t} & =\cfrac{F_{n(n)}}{\delta n}-\cfrac{F_{n(n+\delta n)}}{\delta n}
    -\cfrac{F_{n(n+\delta n)}}{r}+\cfrac{F_{s(s)}-F_{s(s+\delta s)}}{\delta s}

    & =-\cfrac{\partial F_n}{\partial n}-\cfrac{1}{r}F_{n(r+\delta n)}-\cfrac{\partial F_s}{\partial s}

    & =\cfrac{\partial}{\partial s}\left(D_s \cfrac{\partial c}{\partial s}\right)
    +\cfrac{\partial}{\partial n}\left(D_n\cfrac{\partial c}{\partial n}\right)
    +\cfrac{D_n}{r}\cfrac{\partial c}{\partial n}

When :math:`D_s=D_n=D`,

.. math::
    :label: sn_dif6

    \cfrac{\partial c}{\partial t}=D \left(
    \cfrac{\partial^2 c}{\partial s^2}+\cfrac{\partial^2 c}{\partial n^2}+
    \cfrac{1}{r}\cfrac{\partial c}{\partial n}
    \right)


.. _sn_con:

Derivation of the equation of continuity on the s-n coordinates (alternative solution)
==============================================================================================

Let us assume that the :math:`s-n` coordinates are Cartesian curvilinear coordinates. On the :math:`s-n` coordinates, the :math:`s` axis is assumed to be an arbitrary curve, and the :math:`n` axis is assumed to be a straight coordinate axis perpendicular to the `s` axis.

.. _con_sn:

.. figure:: images/0C/con_sn.png
    :width: 80%

    : Flux balance of the infinitesimal elements on the :math:`s-n` coordinates 

:math:`q_s, q_n` is the flow flux. Considering the flow flux balance of the infinitesimal elements in an infinitesimal time shown in :numref:`con_sn`, and taking into account that the area of the infinitesimal element is :math:`r\, \delta\theta\, \delta n`, following equations are derived. 

.. math::
    :label: sn_con1

    \cfrac{\partial h}{\partial t} r\, \delta\theta\, \delta n 
    =q_{n(n)}r \delta\theta - q_{n(n+\delta n)} (r+\delta n) \delta\theta
    +q_{s(s)}\delta n -q_{s(s+\delta s)}\delta n

.. math::
    :label: sn_con2

    \cfrac{\partial h}{\partial t}=\cfrac{q_{n(n)}}{\delta n}-\cfrac{q_{n(n+\delta n)}\, (r+\delta n)}{r\delta n}
    +\cfrac{q_{s(s)}-q_{s(s+\delta s)}}{\delta s}

Where, :math:`h` is the mean water depth for the infinitesimal elements. Then,

.. math::
    :label: sn_con3

    \cfrac{(r+\delta n)}{r\delta n}=\cfrac{1}{r}+\cfrac{1}{\delta n}

Therefore, the second term of the right side of Equation :eq:`sn_con2` is... 

.. math::
    :label: sn_con4

    \cfrac{q_{n(n+\delta n)}\, (r+\delta n)}{r\delta n}=
    \cfrac{q_{n(n+\delta n)}}{r}+\cfrac{q_{n(r+\delta n)}}{\delta n}

Then, Equation :eq:`sn_con2` is expressed as follows.

.. math::
    :label: sn_con5

    \cfrac{\partial h}{\partial t} & =\cfrac{q_{n(n)}}{\delta n}-\cfrac{q_{n(n+\delta n)}}{\delta n}
    -\cfrac{q_{n(n+\delta n)}}{r}+\cfrac{q_{s(s)}-q_{s(s+\delta s)}}{\delta s}

    & =-\cfrac{\partial q_n}{\partial n}-\cfrac{1}{r}q_{n(r+\delta n)}-\cfrac{\partial q_s}{\partial s}

Where,

.. math::
    :label: an_con6

    \cfrac{\partial(rq_n)}{\partial n}=r\cfrac{\partial q_n}{\partial n}+q_n\cfrac{\partial r}{\partial n}=r\cfrac{\partial q_n}{\partial n}+q_n

Consequently,

.. math::
    :label: an_con7

    \cfrac{1}{r}\cfrac{\partial(rq_n)}{\partial n}=\cfrac{\partial q_n}{\partial n}+\cfrac{q_n}{r}

Therefore,

.. math::
    :label: an_con8

    \cfrac{\partial h}{\partial t}+\cfrac{\partial q_s}{\partial s}+\cfrac{1}{r}\cfrac{\partial(rq_n)}{\partial n}=0

or,

.. math::
    :label: an_con9

    \cfrac{\partial h}{\partial t}+\cfrac{\partial(hu_s)}{\partial s}+\cfrac{1}{r}\cfrac{\partial(rhu_n)}{\partial n}=0