<commit_before>.. index::
    triple: compressible; Euler; equations
    single: fluid dynamics

Compressible Euler Equations
============================

The compressible Euler equations are:

.. math::

    {\partial\rho\over\partial t} + \nabla\cdot(\rho{\bf u}) = 0

    {\partial(\rho{\bf u})\over\partial t} + \nabla\cdot(\rho{\bf u}{\bf u}^T)
        + \nabla p - {\bf f} = 0

    {\partial E\over\partial t} + \nabla\cdot({\bf u}(E+p)) = 0

where

.. math::

    E = \rho e + \half \rho u^2

is the total energy per unit volume ($\half \rho u^2$ is the kinetic energy per
unit volume), $e$ is the internal energy per unit mass ($e={U\over nM}$)
and we use the ideal gas equations, so:

.. math::

    e = T c_v

    p = {n\over V} \bar RT =
    {n M\over V} {\bar R\over M}T =
    \rho RT = \rho R {e\over c_v} =
        {R\over c_v} (E-\half \rho u^2)

where
$n$ is the number of moles of gas,
$M$ is the molar mass of the gas (e.g. a mass of one mole of the gas),
$\rho={n M\over V}$ is the density of the gas,
$\bar R$ is the ideal gas constant,
$R={\bar R\over M}$ is the specific ideal gas constant,
$c_v$ is the specific heat capacity at constant volume (e.g. a heat capacity
per unit volume),
$V$ is the volume
and $T$ is the temperature of the gas.
Of those, $V$, $n$, $M$, $R$, $\bar R$ are constants, $\rho$, $e$, $E$ and $T$ are
functions of $(t, x, y, z)$.

We use the substitution:

.. math::

    {\bf U} = \rho {\bf u}

    p = {R\over c_v} (E-\half \rho u^2) =
        {R\over c_v} \left(E-{{\bf U}^2\over2\rho}\right)

and we get:

.. math::

    {\partial\rho\over\partial t} + \nabla\cdot{\bf U} = 0

    {\partial{\bf U}\over\partial t}
        + \nabla\cdot\left({{\bf U}{\bf U}^T\over\rho}+p\one\right)
        - {\bf f} = 0

    {\partial E\over\partial t}
        + \nabla\cdot\left({{\bf U}\over\rho}(E+p)\right) = 0

Now we write ${\bf U} = (U, V, W)$ and we get:

.. math::

       \frac{\partial}{\partial t} \left( \begin{array}{c}
           \varrho\\ U\\ V\\ W\\ E
       \end{array} \right)
       + \frac{\partial}{\partial x} \left( \begin{array}{c}
           U\\
           \frac{U^2}{\varrho} + p\\
           \frac{UV}{\varrho}\\
           \frac{UW}{\varrho}\\
           \frac{U}{\varrho}(E+p)
       \end{array} \right)
       + \frac{\partial}{\partial y} \left( \begin{array}{c}
           V\\
           \frac{VU}{\varrho}\\
           \frac{V^2}{\varrho} + p\\
           \frac{VW}{\varrho}\\
           \frac{V}{\varrho}(E+p)
       \end{array} \right)
       + \frac{\partial}{\partial z} \left( \begin{array}{c}
           W\\
           \frac{WU}{\varrho}\\
           \frac{WV}{\varrho}\\
           \frac{W^2}{\varrho} + p\\
           \frac{W}{\varrho}(E+p)
       \end{array} \right) + \left( \begin{array}{c}
           0\\
           -f_x\\
           -f_y\\
           -f_z\\
           0\\
       \end{array} \right) =
       \left( \begin{array}{c} 0\\ 0\\ 0\\ 0\\ 0 \end{array} \right)

    p = {R\over c_v} \left(E-{U^2+V^2+W^2\over2\rho}\right)

We solve for the unknowns $\rho$, $U$, $V$, $W$ and $E$ as functions of $(t,
x, y, z)$, the rest ($R$, $c_v$, $f_x$, $f_y$, $f_z$) are either constants or
depend on the unknowns.

After introducing:

.. math::

    {\bf w} =
       \left( \begin{array}{c}
           \varrho\\ U\\ V\\ W\\ E
       \end{array} \right)
       =
       \left( \begin{array}{c}
           w_1 \\
           w_2 \\
           w_3 \\
           w_4 \\
           w_5 \\
       \end{array} \right)

    {\bf f}_x =
       \left( \begin{array}{c}
           U\\
           \frac{U^2}{\varrho} + p\\
           \frac{UV}{\varrho}\\
           \frac{UW}{\varrho}\\
           \frac{U}{\varrho}(E+p)
       \end{array} \right)
       =
       \left( \begin{array}{c}
           w_2\\
           \frac{w_2^2}{w_1} + p\\
           \frac{w_2w_3}{w_1}\\
           \frac{w_2w_4}{w_1}\\
           \frac{w_2}{w_1}(w_5+p)
       \end{array} \right)

    {\bf f}_y =
       \left( \begin{array}{c}
           V\\
           \frac{VU}{\varrho}\\
           \frac{V^2}{\varrho} + p\\
           \frac{VW}{\varrho}\\
           \frac{V}{\varrho}(E+p)
       \end{array} \right)
       =
       \left( \begin{array}{c}
           w_3\\
           \frac{w_3w_2}{w_1}\\
           \frac{w_3^2}{w_1} + p\\
           \frac{w_3w_4}{w_1}\\
           \frac{w_3}{w_1}(w_5+p)
       \end{array} \right)

    {\bf f}_z =
       \left( \begin{array}{c}
           W\\
           \frac{WU}{\varrho}\\
           \frac{WV}{\varrho}\\
           \frac{W^2}{\varrho} + p\\
           \frac{W}{\varrho}(E+p)
       \end{array} \right)
       =
       \left( \begin{array}{c}
           w_4\\
           \frac{w_4w_2}{w_1}\\
           \frac{w_4w_3}{w_1}\\
           \frac{w_4^2}{w_1} + p\\
           \frac{w_4}{w_1}(w_5+p)
       \end{array} \right)

    {\bf g} =
       \left( \begin{array}{c}
           0\\
           -f_x\\
           -f_y\\
           -f_z\\
           0\\
       \end{array} \right)

we can then write the equations as:

.. math::

    {\partial{\bf w}\over \partial t} +
    {\partial{\bf f}_x\over \partial x} +
    {\partial{\bf f}_y\over \partial y} +
    {\partial{\bf f}_z\over \partial z} +
    {\bf g}= 0

Now we write the spatial derivatives using so called flux Jacobians
${\bf A}_x$,
${\bf A}_y$
and
${\bf A}_z$:

.. math::

    {\partial{\bf f}_x\over \partial x} =
    {\partial{\bf f}_x\over \partial {\bf w}}
    {\partial{\bf w}\over \partial x} \equiv
    {\bf A}_x
    {\partial{\bf w}\over \partial x}

    {\bf A}_x={\bf A}_x({\bf w})\equiv{\partial{\bf f}_x\over \partial {\bf w}}

Similarly for $y$ and $z$, so we get:

.. math::

    {\partial{\bf w}\over \partial t} +
    {\bf A}_x
    {\partial{\bf w}\over \partial x} +
    {\bf A}_y
    {\partial{\bf w}\over \partial y} +
    {\bf A}_z
    {\partial{\bf w}\over \partial z} +
    {\bf g}= 0

One nice thing about these particular
${\bf f}_x$,
${\bf f}_y$ and
${\bf f}_z$ functions is that they are homogeneous of degree 1:

.. math::

    {\bf f}_x(\lambda{\bf w})
    =\lambda\,{\bf f}_x({\bf w})

so the Euler equation/formula for the homogeneous function is:

.. math::

    {\bf w}\cdot {\partial {\bf f}_x({\bf w})\over\partial {\bf w}}
    ={\bf f}_x({\bf w})

    {\bf w}\cdot {\bf A}_x ={\bf f}_x({\bf w})

So both the ${\bf f}_x$ and it's derivative can be nicely factored out using
the flux Jacobian:

.. math::

    {\bf f}_x = {\bf A}_x\, {\bf w}

    {\partial{\bf f}_x\over \partial x} =
        {\bf A}_x {\partial{\bf w}\over \partial x}

similarly for $y$ and $z$.
To calculate the Jacobians, we'll need:

.. math::

    {\partial p\over \partial {\bf w}}=
        {R\over c_v}
        \left( \begin{array}{ccccc}
            {U^2+V^2+W^2\over 2\rho^2} & -{U\over\rho} & -{V\over\rho}
                & -{W\over\rho} & 1\\
        \end{array} \right)

then we can calculate the Jacobians (and we substitute for $p$):

.. math::

    {\bf J}_x({\bf w}) = {\partial{\bf f}_x\over \partial {\bf w}}=
        \left( \begin{array}{ccccc}
            0 & 1 & 0 & 0 & 0\\
            -{U^2\over\rho^2} +{R\over c_v}{U^2+V^2+W^2\over 2\rho^2} &
                {2U\over\rho}-{R\over c_v}{U\over\rho} &
                -{R\over c_v}{V\over\rho} &
                -{R\over c_v}{W\over\rho} &
                {R\over c_v}\\
            -{UV\over\rho^2} & {V\over\rho} & {U\over\rho} & 0 & 0\\
            -{UW\over\rho^2} & {W\over\rho} & 0 & {U\over\rho} & 0 \\
                -{UE\over\rho^2}-{U\over\rho^2}{R\over c_v}
                    \left(E-{U^2+V^2+W^2\over 2\rho}\right)
                    +{U\over\rho}{R\over c_v}{U^2+V^2+W^2\over 2\rho} &
                {E\over\rho}+{1\over\rho}{R\over c_v}
                    \left(E-{U^2+V^2+W^2\over 2\rho}\right)
                    -{R\over c_v}{U^2\over\rho^2} &
                -{R\over c_v}{UV\over\rho^2} &
                -{R\over c_v}{UW\over\rho^2} &
                {U\over\rho}-{R\over c_v}{U\over\rho} \\
       \end{array} \right)

    {\bf J}_y({\bf w}) = {\partial{\bf f}_y\over \partial {\bf w}}=
        \left( \begin{array}{ccccc}
            0 & 0 & 1 & 0 & 0\\
            -{VU\over\rho^2} & {V\over\rho} & {U\over\rho} & 0 & 0\\
            -{V^2\over\rho^2} +{R\over c_v}{U^2+V^2+W^2\over 2\rho^2} &
                -{R\over c_v}{U\over\rho} &
                {2V\over\rho}-{R\over c_v}{V\over\rho} &
                -{R\over c_v}{W\over\rho} &
                {R\over c_v}\\
            -{VW\over\rho^2} & 0 & {W\over\rho} & {V\over\rho} & 0 \\
            \cdot & \cdot & \cdot & \cdot & \cdot \\
       \end{array} \right)

    {\bf J}_z({\bf w}) = {\partial{\bf f}_z\over \partial {\bf w}}=
        \left( \begin{array}{ccccc}
            0 & 0 & 0 & 1 & 0\\
            -{WU\over\rho^2} & {W\over\rho} & 0 & {U\over\rho} & 0 \\
            -{WV\over\rho^2} & 0 & {W\over\rho} & {V\over\rho} & 0 \\
            -{W^2\over\rho^2} +{R\over c_v}{U^2+V^2+W^2\over 2\rho^2} &
                -{R\over c_v}{U\over\rho} &
                -{R\over c_v}{V\over\rho} &
                {2W\over\rho} -{R\over c_v}{W\over\rho} &
                {R\over c_v}\\
            \cdot & \cdot & \cdot & \cdot & \cdot \\
       \end{array} \right)

Sea Breeze Modeling
-------------------

In our model we make the following assumptions:

.. math::

    f_x = 0

    f_y = 0

    f_z = -\rho g

    V = 0

    {\partial U\over\partial y}
    ={\partial V\over\partial y}
    ={\partial W\over\partial y}
    ={\partial E\over\partial y}=0

so we get:

.. math::

       \frac{\partial}{\partial t} \left( \begin{array}{c}
           \varrho\\ U\\ 0\\ W\\ E<commit_msg>Use the w_i coordinates<commit_after>.. index::
    triple: compressible; Euler; equations
    single: fluid dynamics

Compressible Euler Equations
============================

The compressible Euler equations are:

.. math::

    {\partial\rho\over\partial t} + \nabla\cdot(\rho{\bf u}) = 0

    {\partial(\rho{\bf u})\over\partial t} + \nabla\cdot(\rho{\bf u}{\bf u}^T)
        + \nabla p - {\bf f} = 0

    {\partial E\over\partial t} + \nabla\cdot({\bf u}(E+p)) = 0

where

.. math::

    E = \rho e + \half \rho u^2

is the total energy per unit volume ($\half \rho u^2$ is the kinetic energy per
unit volume), $e$ is the internal energy per unit mass ($e={U\over nM}$)
and we use the ideal gas equations, so:

.. math::

    e = T c_v

    p = {n\over V} \bar RT =
    {n M\over V} {\bar R\over M}T =
    \rho RT = \rho R {e\over c_v} =
        {R\over c_v} (E-\half \rho u^2)

where
$n$ is the number of moles of gas,
$M$ is the molar mass of the gas (e.g. a mass of one mole of the gas),
$\rho={n M\over V}$ is the density of the gas,
$\bar R$ is the ideal gas constant,
$R={\bar R\over M}$ is the specific ideal gas constant,
$c_v$ is the specific heat capacity at constant volume (e.g. a heat capacity
per unit volume),
$V$ is the volume
and $T$ is the temperature of the gas.
Of those, $V$, $n$, $M$, $R$, $\bar R$ are constants, $\rho$, $e$, $E$ and $T$ are
functions of $(t, x, y, z)$.

We use the substitution:

.. math::

    {\bf U} = \rho {\bf u}

    p = {R\over c_v} (E-\half \rho u^2) =
        {R\over c_v} \left(E-{{\bf U}^2\over2\rho}\right)

and we get:

.. math::

    {\partial\rho\over\partial t} + \nabla\cdot{\bf U} = 0

    {\partial{\bf U}\over\partial t}
        + \nabla\cdot\left({{\bf U}{\bf U}^T\over\rho}+p\one\right)
        - {\bf f} = 0

    {\partial E\over\partial t}
        + \nabla\cdot\left({{\bf U}\over\rho}(E+p)\right) = 0

Now we write ${\bf U} = (U, V, W)$ and we get:

.. math::

       \frac{\partial}{\partial t} \left( \begin{array}{c}
           \varrho\\ U\\ V\\ W\\ E
       \end{array} \right)
       + \frac{\partial}{\partial x} \left( \begin{array}{c}
           U\\
           \frac{U^2}{\varrho} + p\\
           \frac{UV}{\varrho}\\
           \frac{UW}{\varrho}\\
           \frac{U}{\varrho}(E+p)
       \end{array} \right)
       + \frac{\partial}{\partial y} \left( \begin{array}{c}
           V\\
           \frac{VU}{\varrho}\\
           \frac{V^2}{\varrho} + p\\
           \frac{VW}{\varrho}\\
           \frac{V}{\varrho}(E+p)
       \end{array} \right)
       + \frac{\partial}{\partial z} \left( \begin{array}{c}
           W\\
           \frac{WU}{\varrho}\\
           \frac{WV}{\varrho}\\
           \frac{W^2}{\varrho} + p\\
           \frac{W}{\varrho}(E+p)
       \end{array} \right) + \left( \begin{array}{c}
           0\\
           -f_x\\
           -f_y\\
           -f_z\\
           0\\
       \end{array} \right) =
       \left( \begin{array}{c} 0\\ 0\\ 0\\ 0\\ 0 \end{array} \right)

    p = {R\over c_v} \left(E-{U^2+V^2+W^2\over2\rho}\right)

We solve for the unknowns $\rho$, $U$, $V$, $W$ and $E$ as functions of $(t,
x, y, z)$, the rest ($R$, $c_v$, $f_x$, $f_y$, $f_z$) are either constants or
depend on the unknowns.

After introducing:

.. math::

    {\bf w} =
       \left( \begin{array}{c}
           \varrho\\ U\\ V\\ W\\ E
       \end{array} \right)
       =
       \left( \begin{array}{c}
           w_1 \\
           w_2 \\
           w_3 \\
           w_4 \\
           w_5 \\
       \end{array} \right)

    {\bf f}_x =
       \left( \begin{array}{c}
           U\\
           \frac{U^2}{\varrho} + p\\
           \frac{UV}{\varrho}\\
           \frac{UW}{\varrho}\\
           \frac{U}{\varrho}(E+p)
       \end{array} \right)
       =
       \left( \begin{array}{c}
           w_2\\
           \frac{w_2^2}{w_1} + p\\
           \frac{w_2w_3}{w_1}\\
           \frac{w_2w_4}{w_1}\\
           \frac{w_2}{w_1}(w_5+p)
       \end{array} \right)

    {\bf f}_y =
       \left( \begin{array}{c}
           V\\
           \frac{VU}{\varrho}\\
           \frac{V^2}{\varrho} + p\\
           \frac{VW}{\varrho}\\
           \frac{V}{\varrho}(E+p)
       \end{array} \right)
       =
       \left( \begin{array}{c}
           w_3\\
           \frac{w_3w_2}{w_1}\\
           \frac{w_3^2}{w_1} + p\\
           \frac{w_3w_4}{w_1}\\
           \frac{w_3}{w_1}(w_5+p)
       \end{array} \right)

    {\bf f}_z =
       \left( \begin{array}{c}
           W\\
           \frac{WU}{\varrho}\\
           \frac{WV}{\varrho}\\
           \frac{W^2}{\varrho} + p\\
           \frac{W}{\varrho}(E+p)
       \end{array} \right)
       =
       \left( \begin{array}{c}
           w_4\\
           \frac{w_4w_2}{w_1}\\
           \frac{w_4w_3}{w_1}\\
           \frac{w_4^2}{w_1} + p\\
           \frac{w_4}{w_1}(w_5+p)
       \end{array} \right)

    {\bf g} =
       \left( \begin{array}{c}
           0\\
           -f_x\\
           -f_y\\
           -f_z\\
           0\\
       \end{array} \right)

    p = {R\over c_v} \left(E-{U^2+V^2+W^2\over2\rho}\right)
    = {R\over c_v} \left(w_5-{w_2^2+w_3^2+w_4^2\over2w_1}\right)

we can then write the equations as:

.. math::

    {\partial{\bf w}\over \partial t} +
    {\partial{\bf f}_x\over \partial x} +
    {\partial{\bf f}_y\over \partial y} +
    {\partial{\bf f}_z\over \partial z} +
    {\bf g}= 0

Now we write the spatial derivatives using so called flux Jacobians
${\bf A}_x$,
${\bf A}_y$
and
${\bf A}_z$:

.. math::

    {\partial{\bf f}_x\over \partial x} =
    {\partial{\bf f}_x\over \partial {\bf w}}
    {\partial{\bf w}\over \partial x} \equiv
    {\bf A}_x
    {\partial{\bf w}\over \partial x}

    {\bf A}_x={\bf A}_x({\bf w})\equiv{\partial{\bf f}_x\over \partial {\bf w}}

Similarly for $y$ and $z$, so we get:

.. math::

    {\partial{\bf w}\over \partial t} +
    {\bf A}_x
    {\partial{\bf w}\over \partial x} +
    {\bf A}_y
    {\partial{\bf w}\over \partial y} +
    {\bf A}_z
    {\partial{\bf w}\over \partial z} +
    {\bf g}= 0

One nice thing about these particular
${\bf f}_x$,
${\bf f}_y$ and
${\bf f}_z$ functions is that they are homogeneous of degree 1:

.. math::

    {\bf f}_x(\lambda{\bf w})
    =\lambda\,{\bf f}_x({\bf w})

so the Euler equation/formula for the homogeneous function is:

.. math::

    {\bf w}\cdot {\partial {\bf f}_x({\bf w})\over\partial {\bf w}}
    ={\bf f}_x({\bf w})

    {\bf w}\cdot {\bf A}_x ={\bf f}_x({\bf w})

So both the ${\bf f}_x$ and it's derivative can be nicely factored out using
the flux Jacobian:

.. math::

    {\bf f}_x = {\bf A}_x\, {\bf w}

    {\partial{\bf f}_x\over \partial x} =
        {\bf A}_x {\partial{\bf w}\over \partial x}

similarly for $y$ and $z$.
To calculate the Jacobians, we'll need:

.. math::

    {\partial p\over \partial {\bf w}}=
        {R\over c_v}
        \left( \begin{array}{ccccc}
            {w_2^2+w_3^2+w_4^2\over 2w_1^2} & -{w_2\over w_1} & -{w_3\over w_1}
                & -{w_4\over w_1} & 1\\
        \end{array} \right)

then we can calculate the Jacobians (and we substitute for $p$):

.. math::

    {\bf J}_x({\bf w}) = {\partial{\bf f}_x\over \partial {\bf w}}=
        \left( \begin{array}{ccccc}
            0 & 1 & 0 & 0 & 0\\
            -{w_2^2\over w_1^2} +{R\over c_v}{w_2^2+w_3^2+w_4^2\over 2 w_1^2} &
                {2w_2\over w_1}-{R\over c_v}{w_2\over w_1} &
                -{R\over c_v}{w_3\over w_1} &
                -{R\over c_v}{w_4\over w_1} &
                {R\over c_v}\\
            -{w_2w_3\over w_1^2} & {w_3\over w_1} & {w_2\over w_1} & 0 & 0\\
            -{w_2w_4\over w_1^2} & {w_4\over w_1} & 0 & {w_2\over w_1} & 0 \\
                -{w_2w_5\over w_1^2}-{w_2\over w_1^2}{R\over c_v}
                    \left(w_5-{w_2^2+w_3^2+w_4^2\over 2 w_1}\right)
                    +{w_2\over w_1}{R\over c_v}{w_2^2+w_3^2+w_4^2\over 2 w_1} &
                {w_5\over w_1}+{1\over w_1}{R\over c_v}
                    \left(w_5-{w_2^2+w_3^2+w_4^2\over 2 w_1}\right)
                    -{R\over c_v}{w_2^2\over w_1^2} &
                -{R\over c_v}{w_2w_3\over w_1^2} &
                -{R\over c_v}{w_2w_4\over w_1^2} &
                {w_2\over w_1}-{R\over c_v}{w_2\over w_1} \\
       \end{array} \right)

    {\bf J}_y({\bf w}) = {\partial{\bf f}_y\over \partial {\bf w}}=
        \left( \begin{array}{ccccc}
            0 & 0 & 1 & 0 & 0\\
            -{w_3w_2\over w_1^2} & {w_3\over w_1} & {w_2\over w_1} & 0 & 0\\
            -{w_3^2\over w_1^2} +{R\over c_v}{w_2^2+w_3^2+w_4^2\over 2 w_1^2} &
                -{R\over c_v}{w_2\over w_1} &
                {2w_3\over w_1}-{R\over c_v}{w_3\over w_1} &
                -{R\over c_v}{w_4\over w_1} &
                {R\over c_v}\\
            -{w_3w_4\over w_1^2} & 0 & {w_4\over w_1} & {w_3\over w_1} & 0 \\
            \cdot & \cdot & \cdot & \cdot & \cdot \\
       \end{array} \right)

    {\bf J}_z({\bf w}) = {\partial{\bf f}_z\over \partial {\bf w}}=
        \left( \begin{array}{ccccc}
            0 & 0 & 0 & 1 & 0\\
            -{w_4w_2\over w_1^2} & {w_4\over w_1} & 0 & {w_2\over w_1} & 0 \\
            -{w_4w_3\over w_1^2} & 0 & {w_4\over w_1} & {w_3\over w_1} & 0 \\
            -{w_4^2\over w_1^2} +{R\over c_v}{w_2^2+w_3^2+w_4^2\over 2 w_1^2} &
                -{R\over c_v}{w_2\over w_1} &
                -{R\over c_v}{w_3\over w_1} &
                {2w_4\over w_1} -{R\over c_v}{w_4\over w_1} &
                {R\over c_v}\\
            \cdot & \cdot & \cdot & \cdot & \cdot \\
       \end{array} \right)

Sea Breeze Modeling
-------------------

In our model we make the following assumptions:

.. math::

    f_x = 0

    f_y = 0

    f_z = -\rho g

    V = 0

    {\partial U\over\partial y}
    ={\partial V\over\partial y}
    ={\partial W\over\partial y}
    ={\partial E\over\partial y}=0

so we get:

.. math::

       \frac{\partial}{\partial t} \left( \begin{array}{c}
           \varrho\\ U\\ 0\\ W\\ E
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{'dir': 'git-commits-cleaned', 'id': '2734509', 'n_tokens_mistral': 8129, 'n_tokens_neox': 7726, 'n_words': 1863}