Editing Apps/RiemannEllipsoidsCompressible (section)

==A Related Two-dimensional Treatment==
===Preamble===
In a related work, [http://adsabs.harvard.edu/abs/1996ApJS..105..181K Korycansky &amp; Papaloizou] (1996, ApJS, 105, 181; hereafter KP96) developed a method to find nontrivial, nonaxisymmetric steady-state flows in a <i>two-dimensional</i> setting.  Specifically, they constructed infinitesimally thin steady-state disk structures in the presence of a time-independent, nonaxisymmetric perturbing potential.  While their problem was only two-dimensional and they did not seek a self-consistent solution of the gravitational Poisson equation, the approach they took to solving the 2D Euler equation in tandem with the continuity equation for a <i>compressible</i> fluid may very well be instructive.  What follows is a summary of their approach.

===Governing Steady-State Equations===

As in our above [http://www.vistrails.org/index.php/User:Tohline/PGE/RotatingFrame#Example_of_Riemann_S-Type_Ellipsoids preamble to discussion of Riemann S-Type Ellipsoids], KP96 set <math>{\vec{\omega}} = \hat{k}\omega</math>.  Hence, their steady-state Euler equation and steady-state continuity equation become (see their Eq. 1 or their Eq. 7),
<div align="center">
<math>
(\vec{v}\cdot \nabla)\vec{v}  + 2\omega\hat{k}\times\vec{v} + \nabla \biggl[H + \Phi -\frac{1}{2}\omega^2 R^2  \biggr] = 0 ,
</math>

<math>
\nabla\cdot(\rho \vec{v}) = \vec{v}\cdot\nabla\rho + \rho\nabla\cdot\vec{v} = 0 .
</math>
</div>
Note that the KP96 notation is slightly different from ours:
* <math>\Sigma</math> is used in place of <math>\rho</math> to denote a two-dimensional <i>surface</i> density;
* <math>\Omega</math> is used instead of <math>\omega</math> to denote the angular frequency of the rotating reference frame;
* <math>W</math> is used instead of <math>H</math> to denote the fluid enthalpy; and
* <math>\Phi_g</math> represents the combined, time-independent gravitational and centrifugal potential, that is, <math>\Phi_g = (\Phi - \omega^2 R^2/2)</math>.

Using the <font color="darkgreen">vector identity</font>, 
<div align="center">
<math>
(\vec{v}\cdot \nabla)\vec{v} = \frac{1}{2}\nabla(v^2) - \vec{v}\times(\nabla\times\vec{v}) ,
</math>
</div>
the above steady-state Euler equation can also be written as,
<div align="center">
<math>
2\omega\hat{k}\times\vec{v} - \vec{v}\times(\nabla\times\vec{v}) + \nabla \biggl[\frac{1}{2}v^2 + H + \Phi -\frac{1}{2}\omega^2 R^2  \biggr] = 0 .
</math>
</div>
Up to this point, no assumptions have been made regarding the behavior of the vector flow-field; we have only chosen to align the <math>\vec{\omega}</math> with the coordinate unit vector, <math>\hat{k}</math>.  In particular, these derived forms for the steady-state Euler and continuity equations can serve to describe a fully 3D problem.

Before proceeding further we should emphasize that, in the context of the Euler equation written in this form (i.e., the form preferred by KP96), the vector <math>\vec{A}</math> defined in the preamble, above, should be written,
<div align="center">
<math>
\vec{A} = 2\omega\hat{k}\times\vec{v} +(\nabla\times\vec{v})\times\vec{v} + \nabla \biggl[\frac{1}{2}v^2 \biggr] .
</math>
</div>


===No Vertical Motions===
Now we restrict the flow by setting <math>v_z = 0</math>, that is, from here on we will assume that all the motion is planar. Also, following the lead of KP96, we define the vorticity of the fluid,
<div align="center">
<math>
\vec{\zeta} \equiv \nabla\times\vec{v} = \hat{i}\zeta_x + \hat{j}\zeta_y + \hat{k}\zeta_z .
</math>
</div> 
[Note that (unfortunately) KP96 use <math>\omega</math> instead of <math>\zeta</math> to represent the rotating-frame vorticity.]  In terms of the components of the vorticity vector, the steady-state Euler equation therefore becomes,
<div align="center">
<math>
(2\omega + \zeta_z)\hat{k}\times\vec{v} + (\hat{i}\zeta_x + \hat{j}\zeta_y)\times\vec{v} + \nabla \biggl[\frac{1}{2}v^2 + H + \Phi -\frac{1}{2}\omega^2 R^2  \biggr] = 0 .
</math>
</div>
Continuing to follow the lead KP96, we next <font color="red">take the curl of this Euler equation</font>.  Because the curl of a gradient is always zero, this leads us to the same condition discussed above &#8212; but this time written in terms of the components of the vorticity &#8212; namely,
<div align="center">
<math>
\nabla\times\vec{A} = 0 = \nabla\times [(2\omega + \zeta_z)\hat{k}\times\vec{v} + (\hat{i}\zeta_x + \hat{j}\zeta_y)\times\vec{v}]  .
</math>
</div>
Using another <font color="darkgreen">vector identity</font>, namely,
<div align="center">
<math>
\nabla\times(\vec{C} \times \vec{B}) =  (\vec{B}\cdot\nabla)\vec{C} - (\vec{C}\cdot\nabla)\vec{B} + \vec{C}(\nabla\cdot\vec{B}) - \vec{B}(\nabla\cdot\vec{C}),
</math>
</div>
and remembering that we are assuming <math>v_z = 0</math>, we see in this case that the vector condition <math>\nabla\times\vec{A}=0</math> leads to the following three independent scalar constraints:
<div align="left">
<math>
~~~~~\hat{i}:~~~~~ [\nabla\times\vec{A}]_x = - \frac{\partial }{\partial z}\biggl[ (2\omega + \zeta)v_x \biggr] + \frac{\partial}{\partial y} \biggl[ \zeta_x v_y - \zeta_y v_x \biggr] = 0  ;
</math><br />
<math>
~~~~~\hat{j}:~~~~~ [\nabla\times\vec{A}]_y = - \frac{\partial }{\partial z} \biggl[ (2\omega + \zeta)v_y  \biggr] - \frac{\partial}{\partial x} \biggl[ \zeta_x v_y - \zeta_y v_x \biggr] = 0 ;
</math><br />
<math>
~~~~~\hat{k}:~~~~~ [\nabla\times\vec{A}]_z = (2\omega + \zeta)\nabla\cdot\vec{v} + \biggl[ v_x \frac{\partial}{\partial x} + v_y \frac{\partial}{\partial y} \biggr](2\omega + \zeta) = 0 .
</math>
</div>
With the understanding that, by definition,
<div align="center">
<math>
\zeta_x \equiv - \frac{\partial v_y}{\partial z} , ~~~~~
\zeta_y \equiv + \frac{\partial v_x}{\partial z} ,  ~~~~~ \mathrm{and} ~~~~~
\zeta_z \equiv \frac{\partial v_y}{\partial x} - \frac{\partial v_x}{\partial y} ,
</math>
</div>
it can be shown that these three constraints are identical to the ones presented in the preamble, above.

===Solution Strategy===

<font color="darkblue"><b>Constraint #1:</b></font>
For their two-dimensional disk problem, KP96 focused on the constraint provided by the z-component of the curl of the Euler equation, which can be rewritten as (see above derivation, or Eq. 2 of KP96),
<div align="center">
<math>
\nabla\cdot\vec{v} =-\vec{v} \cdot \biggl[ \frac{\nabla(2\omega + \zeta_z)}{(2\omega + \zeta_z)} \biggr] 
= -\vec{v} \cdot \nabla[\ln(2\omega + \zeta_z)].
</math>
</div>
<font color="darkblue"><b>Constraint #2:</b></font>
But from the continuity equation they also know that,
<div align="center">
<math>
\nabla\cdot\vec{v} = -\vec{v}\cdot\biggl[\frac{\nabla\rho}{\rho} \biggr] = -\vec{v} \cdot \nabla[\ln\rho] .
</math>
</div>
Hence,
<div align="center">
<math>
\vec{v} \cdot \nabla[\ln(2\omega + \zeta_z)]  = \vec{v} \cdot \nabla[\ln\rho] ,
</math>
</div>
that is,
<div align="center">
<math>
\vec{v} \cdot \nabla\ln\biggl[ \frac{(2\omega + \zeta_z)}{\rho} \biggr]  = 0  .
</math>
</div>
This is essentially KP96's Eq. (3).

<font color="darkblue"><b>Introduce stream function:</b></font>
The constraint implied by the continuity equation also suggests that it might be useful to define a stream function in terms of the momentum density &#8212; instead of in terms of just the velocity, which is the natural treatment in the context of incompressible fluid flows.  KP96 do this. They define the stream function, <math>\Psi</math>, such that (see their Eq. 4),
<div align="center">
<math>
\rho\vec{v} = \nabla\times(\hat{k}\Psi)  .
</math>
</div>
in which case,
<div align="center">
<math>
v_x = \frac{1}{\rho} \frac{\partial \Psi}{\partial y} ~~~~~\mathrm{and}~~~~~  
v_y = - \frac{1}{\rho} \frac{\partial \Psi}{\partial x} .
</math>
</div>
This implies as well that the z-component of the fluid vorticity can be expressed in terms
of the stream function as follows (see Eq. 5 of KP96):
<div align="center">
<math>
\zeta_z = - \nabla\cdot \biggl( \frac{\nabla\Psi}{\rho} \biggr) = - \frac{\partial}{\partial x} \biggl[ \frac{1}{\rho} \frac{\partial\Psi}{\partial x} \biggr] - \frac{\partial}{\partial y} \biggl[ \frac{1}{\rho} \frac{\partial\Psi}{\partial y} \biggr].
</math>
</div>

According to KP96, this expression, taken in combination with the conclusion drawn above from the second constraint &#8212; that is, Eq. (3) taken in combination with Eq. (4) from KP96 &#8212; "tell us that the 'vortensity' <math>(\zeta_z + 2\omega)/\rho</math> is constant along streamlines which are lines of constant <math>\Psi</math>."  The vortensity is therefore a function of <math>\Psi</math> alone, so we can write,
<div align="center">
<math>
\frac{\zeta_z + 2\omega}{\rho} = g(\Psi) .
</math>
</div>


<font color="darkblue"><b>Constraint #3:</b></font>

Taking the scalar product of <math>\vec{v}</math> and the following form of the steady-state Euler equation, 
<div align="center">
<math>
2\omega\hat{k}\times\vec{v} - \vec{v}\times(\nabla\times\vec{v}) + \nabla \biggl[\frac{1}{2}v^2 + H + \Phi -\frac{1}{2}\omega^2 R^2  \biggr] = 0 ,
</math>
</div>
we obtain the constraint,
<div align="center">
<math>
\vec{v}\cdot\nabla \biggl[\frac{1}{2}v^2 + H + \Phi -\frac{1}{2}\omega^2 R^2  \biggr] = 0 .
</math>
</div>
When tied with our earlier discussion, this means that the Bernoulli function also must be constant along streamlines.  Hence, we can write,
<div align="center">
<math>
\frac{1}{2}v^2 + H + \Phi -\frac{1}{2}\omega^2 R^2  = F(\Psi) .
</math>
</div>
KP96 then go on to demonstrate that the relationship between the functions <math>g(\Psi)</math> and <math>F(\Psi)</math> is,
<div align="center">
<math>
\frac{dF}{d\Psi} = -g(\Psi) ,
</math>
</div>
which allows the determination of <math>F</math> up to a constant of integration.

===Summary===
In summary, KP96 constrain their flow as follows:
# They use the z-component of the curl of the Euler equation;
# They use the compressible version of the continuity equation;
# Instead of taking the divergence of the Euler equation to obtain a Poisson-like equation, they obtain an algebraic constraint on the Bernoulli function (as in our traditional SCF technique) by simply "dotting" <math>\vec{v}</math> into the Euler equation.

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