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==Compare== ===Incompressible=== <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\bold{u}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\lambda \biggl[ \boldsymbol{\hat\imath} \biggl( \frac{a}{b}\biggr) y - \boldsymbol{\hat\jmath} \biggl( \frac{b}{a} \biggr) x \biggr] \, .</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ u^2 \equiv \bold{u}\cdot \bold{u}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \lambda^2 \biggl[ b^2 \biggl(\frac{x^2}{a^2}\biggr) + a^2 \biggl(\frac{y^2}{b^2}\biggr) \biggr] \, . </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ \boldsymbol{\zeta} \equiv \boldsymbol{\nabla \times}\bold{u}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \boldsymbol{\hat\imath} \biggl[ \frac{\partial u_z}{\partial y} - \frac{\partial u_y}{\partial z} \biggr] + \boldsymbol{\hat\jmath} \biggl[ \frac{\partial u_x}{\partial z} - \frac{\partial u_z}{\partial x} \biggr] + \bold{\hat{k}} \biggl[ \frac{\partial u_y}{\partial x} - \frac{\partial u_x}{\partial y} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\bold{\hat{k}} \biggl[ - \lambda \biggl(\frac{b}{a} + \frac{a}{b}\biggr) \biggr] \, .</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ \boldsymbol{\zeta} \times \bold{u} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - \lambda^2 \biggl(\frac{b}{a} + \frac{a}{b}\biggr) \biggl[ \boldsymbol{\hat\imath} \biggl(\frac{b}{a}\biggr)x + \boldsymbol{\hat\jmath} \biggl(\frac{a}{b}\biggr)y \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- \nabla \biggl\{ \frac{\lambda^2(a^2 + b^2)}{2} \biggl[ \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr] \biggr\} \, . </math> </td> </tr> </table> Returning to the above-mentioned, <div align="center"> <font color="#770000">'''Eulerian Representation'''</font><br /> of the Euler Equation <br /> <font color="#770000">'''as viewed from a Rotating Reference Frame'''</font> <math>\frac{\partial \bold{u}}{\partial t} + (\bold{u}\cdot \nabla) \bold{u} = - \frac{1}{\rho} \nabla P - \nabla \Phi_\mathrm{grav} - {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x}) - 2{\vec{\Omega}}_f \times \bold{u} \, ,</math> </div> we next note — as we have done in our [[PGE/Euler#in_terms_of_the_vorticity:|broader discussion of the Euler equation]] — that, <div align="center"> <math> (\bold{u} \cdot\nabla)\bold{u} = \frac{1}{2}\nabla(\bold{u} \cdot \bold{u}) - \bold{u} \times(\nabla\times\bold{u}) = \frac{1}{2}\nabla(u^2) + \boldsymbol\zeta \times \bold{u} \, . </math> </div> Making this substitution, we obtain <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ \frac{\partial \bold{u}}{\partial t} + \biggl[ \frac{1}{2}\nabla(u^2) + \boldsymbol\zeta \times \bold{u} \biggr] </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - \frac{1}{\rho} \nabla P - \nabla \Phi_\mathrm{grav} - {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x}) - 2{\vec{\Omega}}_f \times \bold{u} </math> </td> </tr> </table> it can be shown that, for the velocity fields associated with all Riemann S-type ellipsoids, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~({\vec{v}}_{rot}\cdot \nabla) {\vec{v}}_{rot}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -\nabla \biggl[ \frac{1}{2} \lambda^2(x^2 + y^2) \biggr] \, ; </math> </td> </tr> <tr> <td align="right"> <math>~- {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x})</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ +\nabla\biggl[\frac{1}{2} \Omega_f^2 (x^2 + y^2) \biggr] \, ; </math> </td> </tr> <tr> <td align="right"> <math>~- 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - \nabla\biggl[ \Omega_f \lambda\biggl( \frac{b}{a} x^2 + \frac{a}{b}y^2 \biggr) \biggr] \, . </math> </td> </tr> </table> As a check, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~(\bold{u} \cdot\nabla)\bold{u}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{1}{2}\nabla(u^2) + \boldsymbol\zeta \times \bold{u} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\nabla \biggl\{\frac{\lambda^2}{2} \biggl[ b^2 \biggl(\frac{x^2}{a^2}\biggr) + a^2 \biggl(\frac{y^2}{b^2}\biggr) \biggr] - \frac{\lambda^2(a^2 + b^2)}{2} \biggl[ \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right" colspan="3"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\nabla \frac{\lambda^2}{2} \biggl\{ -(x^2 + y^2) \biggr\} \, . </math> </td> </tr> </table> Yes! This expression matches the one that appears just a few lines earlier in this discussion. Now, let's switch the order of terms in the steady-state Euler equation to permit an easier comparison with our attempt to develop the compressible models. <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ \cancel{\frac{\partial \bold{u}}{\partial t}} + \biggl[ \frac{1}{2}\nabla(u^2) + \boldsymbol\zeta \times \bold{u} \biggr] </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - \frac{1}{\rho} \nabla P - \nabla \Phi_\mathrm{grav} - {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x}) - 2{\vec{\Omega}}_f \times \bold{u} </math> </td> </tr> <tr> <td align="right"> <math>~ \Rightarrow~~~ 0 </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - \nabla \biggl[ H + \Phi_\mathrm{grav} + \frac{1}{2} u^2 - \frac{1}{2} \Omega_f^2 (x^2 + y^2) \biggr] - (\boldsymbol\zeta + 2{\vec{\Omega}}_f )\times \bold{u} \, . </math> </td> </tr> </table> ===Compressible=== Now in our [[#AndalibBernoulli|above discussion of Andalib's work]], the steady-state form of the Euler equation was formulated as, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>\nabla \biggl[H + \Phi_\mathrm{grav} + F_B(\Psi) + \frac{1}{2}u^2 - \frac{1}{2}\Omega_f^2 (x^2 + y^2) \biggr] </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 0 \, .</math> </td> </tr> </table> It is easy to appreciate that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\nabla F_B(\Psi) </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~(\boldsymbol\zeta + 2{\vec{\Omega}}_f )\times \bold{u} \, .</math> </td> </tr> </table> As we have shown, in the case of the incompressible (Riemann S-type ellipsoid) models, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~(\boldsymbol\zeta + 2{\vec{\Omega}}_f )\times \bold{u}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\nabla \biggl\{ \frac{\lambda^2(a^2 + b^2)}{2} \biggl[ \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr] + \biggl[ \Omega_f \lambda\biggl( \frac{b}{a} x^2 + \frac{a}{b}y^2 \biggr) \biggr] \biggr\} \, . </math> </td> </tr> </table> If we attempt to directly relate these two expressions, we must acknowledge that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~F_B(\Psi)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{\lambda^2(a^2 + b^2)}{2} \biggl[ \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr] + \biggl[ \Omega_f \lambda\biggl( \frac{b}{a} x^2 + \frac{a}{b}y^2 \biggr) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{x^2}{a^2} \biggl[\frac{1}{2}\lambda^2(a^2 + b^2) + \Omega_f \lambda a b \biggr] + \frac{y^2}{b^2} \biggl[ \frac{1}{2}\lambda^2(a^2 + b^2) + \Omega_f \lambda a b \biggr] \, . </math> </td> </tr> </table> As we have discussed above, Andalib (1998) found that some interesting model sequences could be constructed if he adopted the functional form, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~F_B(\Psi)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~C_0 \Psi + \frac{1}{2} C_1 \Psi^2 \, .</math> </td> </tr> </table> Evidently, the incompressible (uniform-density) Riemann S-type ellisoids can be retrieved from our derived compressible-model formalism if we set, <math>~C_1 = 0</math>, and, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Psi(x,y)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{x^2}{a^2} + \frac{y^2}{b^2} \, ,</math> </td> </tr> </table> with, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~C_0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{1}{2}\lambda^2(a^2 + b^2) + \Omega_f \lambda a b \biggr] </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- \frac{\zeta_z}{2}\biggl( \frac{a^2 b^2}{a^2 + b^2}\biggr)\biggl[ 2\Omega_f - \zeta_z \biggr] \, .</math> </td> </tr> </table> <font color="red">Afterthought:</font> Because we want <math>~\Psi(x,y)</math> to go to zero at the surface, it likely will be better to set, <div align="center"> <math>~\Psi \equiv 1 - \biggl[ \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr] \, ,</math> </div> then adjust the sign of <math>~C_0</math> and add a constant (zeroth-order term) to the definition of <math>~F_B(\Psi)</math>. ===Trial #1=== Restricting our discussion to nonaxisymmetric, thin disks, let's assume <math>~\rho</math> is uniform throughout the configuration and that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Psi(x,y)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \Psi_0 \biggl[1 - \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) \biggr] \, . </math> </td> </tr> </table> This means that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{ \partial \Psi}{\partial x} = - \Psi_0 \biggl( \frac{2x}{a^2}\biggr)</math> </td> <td align="center"> and, </td> <td align="left"> <math>~\frac{ \partial \Psi}{\partial y} = - \Psi_0 \biggl( \frac{2y}{b^2}\biggr) \, .</math> </td> </tr> </table> The momentum density vector is governed by the relation, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\rho\bold{u}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \boldsymbol{\hat\imath} \biggl[ \frac{\partial \Psi}{\partial y} \biggr] - \boldsymbol{\hat\jmath} \biggl[ \frac{\partial \Psi}{\partial x}\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \Psi_0 \biggl\{- \boldsymbol{\hat\imath} \biggl[ \frac{2y}{b^2} \biggr] + \boldsymbol{\hat\jmath} \biggl[ \frac{2x}{a^2}\biggr] \biggr\} \, . </math> </td> </tr> <tr><td align="center" colspan="3">[<math>~\Psi</math> has units of "density × length<sup>2</sup> per time"]</td> </table> First of all, let's see if the steady-state continuity equation is satisfied: <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\nabla \cdot (\rho \bold{u})</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{\partial (\rho u_x)}{\partial x} + \frac{\partial (\rho u_y)}{\partial y} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\rho \biggl\{ \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\Psi_0 \rho \biggl\{ \frac{\partial }{\partial x}\biggl[ - \frac{2y}{b^2} \biggr] + \frac{\partial }{\partial y}\biggl[ \frac{2x}{a^2} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~0 \, .</math> <font color="red">Q.E.D.</font> </td> </tr> </table> Next, let's determine the z-component of the vorticity and the vortensity: <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\zeta_z</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{\partial u_y}{\partial x} - \frac{\partial u_x}{\partial y} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\Psi_0 \biggl\{ \frac{\partial }{\partial x}\biggl[ \frac{2x}{\rho a^2} \biggr] + \frac{\partial }{\partial y}\biggl[ \frac{2y}{\rho b^2} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{2\Psi_0}{\rho} \biggl[ \frac{1}{a^2} + \frac{1}{b^2} \biggr] \, ; </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ \frac{(\zeta_z + 2\Omega_f)}{\rho}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{1}{\rho} \biggl\{ \frac{2\Psi_0}{\rho} \biggl[ \frac{1}{a^2} + \frac{1}{b^2} \biggr] + 2\Omega_f \biggr\}\, . </math> </td> </tr> </table> This means that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ (\vec\zeta + 2\vec\Omega) \times \bold{u} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl\{ \frac{(\zeta_z + 2\Omega_f)}{\rho} \biggr\} \boldsymbol{\hat{k}} \times (\rho \bold{u} )</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\Psi_0 \biggl\{ \frac{(\zeta_z + 2\Omega_f)}{\rho} \biggr\} \boldsymbol{\hat{k}} \times \biggl\{- \boldsymbol{\hat\imath} \biggl[ \frac{2y}{b^2} \biggr] + \boldsymbol{\hat\jmath} \biggl[ \frac{2x}{a^2}\biggr] \biggr\}</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-2 \Psi_0 \biggl[ \frac{(\zeta_z + 2\Omega_f)}{\rho} \biggr] \biggl[ \boldsymbol{\hat\imath} \biggl( \frac{x}{a^2}\biggr) + \boldsymbol{\hat\jmath} \biggl( \frac{y}{b^2} \biggr) \biggr] \, .</math> </td> </tr> </table> But this entire process was designed to ensure that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ (\vec\zeta + 2\vec\Omega) \times \bold{u} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\nabla F_B(\Psi) \, ,</math> </td> </tr> </table> where, <math>~F_B(\Psi) \equiv C_0 \Psi</math>. Let's see if we get the same expression … <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\nabla F_B(\Psi)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \boldsymbol{\hat\imath} \frac{\partial}{\partial x} + \boldsymbol{\hat\jmath} \frac{\partial}{\partial y} \biggr]C_0 \Psi</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ C_0 \Psi_0\biggl[ \boldsymbol{\hat\imath} \frac{\partial}{\partial x} + \boldsymbol{\hat\jmath} \frac{\partial}{\partial y} \biggr] \biggl[1 - \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - 2C_0 \Psi_0\biggl[ \boldsymbol{\hat\imath} \biggl( \frac{x}{a^2} \biggr) + \boldsymbol{\hat\jmath} \biggl( \frac{y}{b^2} \biggr) \biggr] \, . </math> </td> </tr> </table> This is indeed the same expression as above if we set, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~C_0 </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{(\zeta_z + 2\Omega_f)}{\rho} \, .</math> </td> </tr> </table> <font color="red">Hooray!!</font> Finally, let's make sure that the elliptic PDE "constraint" equation is satisfied. <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ 2\Omega_f </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~C_0 \rho+~C_1 \rho \Psi + \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> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~(\zeta_z + 2\Omega_f)+~\cancel{C_1 \rho \Psi} - \frac{\partial }{\partial x}\biggl[\frac{\Psi_0}{\rho} \frac{2x}{a^2} \biggr] - \frac{\partial }{\partial y} \biggl[\frac{\Psi_0}{\rho} \frac{2y}{b^2} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\zeta_z + 2\Omega_f - \biggl[\frac{\Psi_0}{\rho} \frac{2}{a^2} \biggr] - \biggl[\frac{\Psi_0}{\rho} \frac{2}{b^2} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\zeta_z + 2\Omega_f - \frac{2\Psi_0}{\rho}\biggl[\frac{1}{a^2} + \frac{1}{b^2} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~2\Omega_f \, .</math> <font color="red">Yes!</font> </td> </tr> </table> ===Trial #2=== Still restricting our discussion to nonaxisymmetric, thin disks, let's try, <math>~\Psi = \Psi_0 (\rho/\rho_c)^2</math>, and <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\rho</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\rho_c \biggl\{1 - \biggl[ \frac{y^2}{b^2} + \frac{x^2}{a^2}\biggr]\biggr\} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~\frac{\partial^2 \rho}{\partial x^2} = -\frac{\partial}{\partial x}\biggl\{ \frac{2\rho_c x}{a^2}\biggr\} = - \frac{2\rho_c}{a^2}</math> </td> <td align="center"> and, </td> <td align="left"> <math>~\frac{\partial^2 \rho}{\partial y^2} = - \frac{\partial}{\partial y}\biggl\{ \frac{2\rho_c y}{b^2} \biggr\} = - \frac{2\rho_c}{b^2} \, .</math> </td> </tr> </table> <table border="1" align="center" width="90%" cellpadding="10"><tr><td align="left"> <font color="red">'''IMPORTANT NOTE''' (by Tohline on 22 September 2020):</font> As I have come to appreciate today — after studying the relevant sections of both [[Appendix/References#EFE|EFE]] and [[Appendix/References#BT87|BT87]] — this is an example of a heterogeneous density distribution whose gravitational potential has an analytic prescription. As is discussed in a [[ThreeDimensionalConfigurations/HomogeneousEllipsoids#Inhomogeneous_Ellipsoids_Leading_to_Ferrers_Potentials| separate chapter]], the potential that it generates is sometimes referred to as a ''Ferrers'' potential, for the exponent, n = 1. In our [[ThreeDimensionalConfigurations/HomogeneousEllipsoids#GravFor1|accompanying discussion]] we find that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{ \Phi_\mathrm{grav}(\bold{x})}{(-\pi G\rho_c)} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{2} I_\mathrm{BT} a_1^2 - \biggl(A_1 x^2 + A_2 y^2 +A_3 z^2 \biggr) ~+ \biggl( A_{12} x^2y^2 + A_{13} x^2z^2 + A_{23} y^2z^2\biggr) ~+ \frac{1}{6} \biggl(3A_{11}x^4 + 3A_{22}y^4 + 3A_{33}z^4 \biggr) \, , </math> </td> </tr> </table> where, <table border="0" align="center" cellpadding="10" width="80%"> <tr> <td align="center" width="50%"> <table border="0" cellpadding="5" align="center"> <tr><td align="center" colspan="3">for <math>~i \ne j</math></td></tr> <tr> <td align="right"> <math>~A_{ij}</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~-\frac{A_i-A_j}{(a_i^2 - a_j^2)} </math> </td> </tr> <tr><td align="center" colspan="3">[ [[Appendix/References#EFE|EFE]], <font color="#00CC00">§21, Eq. (107)</font> ]</td></tr> </table> </td> <td align="center" width="50%"> <table border="0" cellpadding="5" align="center"> <tr><td align="center" colspan="3">for <math>~i = j</math></td></tr> <tr> <td align="right"> <math>~2A_{ii} + \sum_{\ell = 1}^3 A_{i\ell}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{2}{a_i} </math> </td> </tr> <tr><td align="center" colspan="3">[ [[Appendix/References#EFE|EFE]], <font color="#00CC00">§21, Eq. (109)</font> ]</td></tr> </table> </td> </tr> </table> More specifically, in the three cases where the indices, <math>~i=j</math>, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~3A_{11}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{2}{a_1^2} - (A_{12} + A_{13}) \, , </math> </td> </tr> <tr> <td align="right"> <math>~3A_{22}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{2}{a_2^2} - (A_{21} + A_{23}) \, , </math> </td> </tr> <tr> <td align="right"> <math>~3A_{33}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{2}{a_3^2} - (A_{31} + A_{32}) \, . </math> </td> </tr> </table> </td></tr></table> This means that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{ \partial \Psi}{\partial x} = \biggl( \frac{\Psi_0}{\rho_c^2}\biggr) 2\rho \frac{ \partial \rho}{\partial x} = -2\rho \biggl( \frac{2\rho_c x}{a^2} \biggr)\biggl( \frac{\Psi_0}{\rho_c^2}\biggr)</math> </td> <td align="center"> and, </td> <td align="left"> <math>~\frac{ \partial \Psi}{\partial y} = \biggl( \frac{\Psi_0}{\rho_c^2}\biggr)2\rho \frac{ \partial \rho}{\partial y} = -2\rho \biggl( \frac{2\rho_c y}{b^2} \biggr)\biggl( \frac{\Psi_0}{\rho_c^2}\biggr) \, .</math> </td> </tr> </table> <span id="ConstraintTrial2"> </span> <table border="1" align="center" cellpadding="10" width="80%"><tr><td align="left"> <div align="center">'''The Elliptic PDE Constraint Equation'''</div> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ 2\Omega_f </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~C_0 \rho+~\cancel{C_1 \rho \Psi} + \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> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~C_0 \rho - \frac{\partial }{\partial x}\biggl[ \biggl( \frac{4 x}{a^2} \biggr)\biggl( \frac{\Psi_0}{\rho_c}\biggr) \biggr] - \frac{\partial }{\partial y} \biggl[ \biggl( \frac{4 y}{b^2} \biggr)\biggl( \frac{\Psi_0}{\rho_c}\biggr) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~C_0 \rho - \frac{4\Psi_0}{\rho_c}\biggl[ \frac{1}{a^2} + \frac{1}{b^2} \biggr] </math> </td> </tr> </table> </td></tr></table> The momentum density vector is governed by the relation, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\rho\bold{u}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \boldsymbol{\hat\imath} \biggl[ \frac{\partial \Psi}{\partial y} \biggr] - \boldsymbol{\hat\jmath} \biggl[ \frac{\partial \Psi}{\partial x}\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - \boldsymbol{\hat\imath} \biggl[ 2\rho \biggl( \frac{2\rho_c y}{b^2} \biggr)\biggl( \frac{\Psi_0}{\rho_c^2}\biggr) \biggr] + \boldsymbol{\hat\jmath} \biggl[ 2\rho \biggl( \frac{2\rho_c x}{a^2} \biggr)\biggl( \frac{\Psi_0}{\rho_c^2}\biggr)\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \rho \biggl\{ - \boldsymbol{\hat\imath} \biggl[ \frac{4\Psi_0 y}{\rho_c b^2} \biggr] + \boldsymbol{\hat\jmath} \biggl[ \frac{4\Psi_0 x}{\rho_c a^2} \biggr] \biggr\} \, . </math> </td> </tr> <tr><td align="center" colspan="3">[<math>~\Psi</math> has units of "density × length<sup>2</sup> per time"]</td> </table> As above, let's see if the steady-state continuity equation is satisfied: <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\nabla \cdot (\rho \bold{u})</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{\partial (\rho u_x)}{\partial x} + \frac{\partial (\rho u_y)}{\partial y} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -\frac{\partial }{\partial x}\biggl[ \frac{4\Psi_0 y \rho}{\rho_c b^2} \biggr] + \frac{\partial }{\partial y}\biggl[ \frac{4\Psi_0 x \rho }{\rho_c a^2} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{4\Psi_0}{\rho_c} \biggl\{ - \frac{y}{b^2} \cdot \frac{\partial \rho}{\partial x} + \frac{ x }{ a^2} \cdot \frac{\partial \rho}{\partial y} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{4\Psi_0}{\rho_c} \biggl\{ \frac{y}{b^2} \biggl[ \frac{2\rho_c x}{a^2} \biggr] - \frac{ x }{ a^2} \biggl[ \frac{2 \rho_c y}{b^2} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~0 \, .</math> <font color="red">Yes!</font> </td> </tr> </table> Next, as above, let's determine the z-component of the vorticity and the vortensity: <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\zeta_z</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{\partial u_y}{\partial x} - \frac{\partial u_x}{\partial y} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{ 4\Psi_0}{\rho_c} \biggl\{ \frac{\partial }{\partial x}\biggl[ \frac{x}{a^2} \biggr] + \frac{\partial }{\partial y}\biggl[ \frac{y}{b^2 } \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{ 4\Psi_0}{\rho_c} \biggl[ \frac{1}{a^2} + \frac{1}{b^2} \biggr] \, . </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ \frac{(\zeta_z + 2\Omega_f)}{\rho}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{1}{\rho} \biggl\{ \frac{ 4\Psi_0}{\rho_c} \biggl[ \frac{1}{a^2} + \frac{1}{b^2} \biggr] + 2\Omega_f \biggr\}\, . </math> </td> </tr> </table> This means that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ (\vec\zeta + 2\vec\Omega) \times \bold{u} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl\{ \frac{(\zeta_z + 2\Omega_f)}{\rho} \biggr\} \boldsymbol{\hat{k}} \times (\rho \bold{u} )</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{ 4\Psi_0 (\zeta_z + 2\Omega_f)}{\rho_c} \boldsymbol{\hat{k}} \times \biggl\{ - \boldsymbol{\hat\imath} \biggl[ \frac{y}{b^2} \biggr] + \boldsymbol{\hat\jmath} \biggl[ \frac{x}{a^2} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- \frac{ 4\Psi_0 (\zeta_z + 2\Omega_f)}{\rho_c} \biggl\{ \boldsymbol{\hat\imath} \biggl[ \frac{x}{a^2} \biggr] + \boldsymbol{\hat\jmath} \biggl[ \frac{y}{b^2} \biggr] \biggr\} </math> </td> </tr> </table> Now, let's examine the gradient of <math>~F_B(\Psi)</math>. <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\nabla F_B(\Psi)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \boldsymbol{\hat\imath} \frac{\partial}{\partial x} + \boldsymbol{\hat\jmath} \frac{\partial}{\partial y} \biggr]C_0 \Psi</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{ C_0 \Psi_0}{\rho_c^2} \biggl[ \boldsymbol{\hat\imath} \frac{\partial \rho^2}{\partial x} + \boldsymbol{\hat\jmath} \frac{\partial \rho^2}{\partial y} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{ 2C_0 \rho \Psi_0}{\rho_c^2} \biggl\{ -\boldsymbol{\hat\imath} \biggl[ \frac{2 \rho_c x}{a^2} \biggr] - \boldsymbol{\hat\jmath} \biggl[ \frac{2 \rho_c y}{b^2} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - \frac{ 4 C_0 \rho \Psi_0}{\rho_c} \biggl\{ \boldsymbol{\hat\imath} \biggl[ \frac{x}{a^2} \biggr] + \boldsymbol{\hat\jmath} \biggl[ \frac{y}{b^2} \biggr] \biggr\} \, , </math> </td> </tr> </table> which is identical to the immediately preceding expression if we set, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~C_0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{(\zeta_z + 2\Omega_f)}{\rho} \, . </math> </td> </tr> </table> Continuing with the [[#ConstraintTrial2|above examination of the elliptic PDE "constraint" equation]], we find that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~2\Omega_f</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~C_0 \rho - \zeta_z </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~(2\Omega_f + \zeta_z) - \zeta_z </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~2\Omega_f \, .</math> <font color="red">Hooray!</font> </td> </tr> </table> ===Trial #3=== If the density distribution has been specified, then <math>~\Psi</math> is the "stream-function" from which all rotating-frame velocities are determined. Specifically, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\bold{u}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{\rho} \biggl[ \boldsymbol{\hat\imath} \frac{\partial \Psi}{\partial y} - \boldsymbol{\hat\jmath} \frac{\partial \Psi}{\partial x} \biggr] \, . </math> </td> </tr> </table> Most importantly, as has been [[#Related_Useful_Expressions|detailed above]], the term on the left-hand-side of the steady-state Euler equation becomes, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~(\vec\zeta + 2\vec\Omega) \times \bold{u}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{ (\zeta_z + 2\Omega_f)}{\rho} \biggr] \nabla\Psi \, ,</math> </td> </tr> <tr><td align="center" colspan="3">[https://digitalcommons.lsu.edu/gradschool_disstheses/6650/ Saied W. Andalib (1998)], §4.2, p. 83, Eq. (4.13)</td></tr> </table> where, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\zeta_z </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~\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> </td> </tr> </table> Still restricting our discussion to infinitesimally thin, nonaxisymmetric disks, let's assume that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\rho</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\rho_c \biggl[1 - \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) \biggr] </math> </td> </tr> </table> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Rightarrow~~~\frac{\partial \rho}{\partial x} = -\biggl( \frac{2\rho_c x}{a^2} \biggr)</math> </td> <td align="center"> and, </td> <td align="left"> <math>~\frac{\partial \rho}{\partial y} = -\biggl( \frac{2\rho_c y}{b^2} \biggr) \, ;</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~\frac{\partial^2 \rho}{\partial x^2} = -\frac{\partial}{\partial x}\biggl\{ \frac{2\rho_c x}{a^2}\biggr\} = - \frac{2\rho_c}{a^2}</math> </td> <td align="center"> and, </td> <td align="left"> <math>~\frac{\partial^2 \rho}{\partial y^2} = - \frac{\partial}{\partial y}\biggl\{ \frac{2\rho_c y}{b^2} \biggr\} = - \frac{2\rho_c}{b^2} \, .</math> </td> </tr> </table> And, let's assume that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Psi</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\Psi_0 \biggl( \frac{\rho}{\rho_c} \biggr)^q \, ,</math> </td> </tr> </table> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Rightarrow~~~ \frac{\partial\Psi}{\partial x} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \rho^{q-1}\biggl(\frac{q \Psi_0}{\rho_c^q}\biggr) \frac{\partial\rho}{\partial x} = - \rho^{q-1}\biggl(\frac{q \Psi_0}{\rho_c^q}\biggr) \biggl(\frac{2\rho_c x}{a^2} \biggr) = - \biggl( \frac{\rho}{\rho_c}\biggr)^{q-1} \biggl(\frac{2q \Psi_0x}{a^2} \biggr) \, ; </math> </td> </tr> <tr> <td align="right"> and, similarly, <math>~\frac{\partial\Psi}{\partial y} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - \biggl( \frac{\rho}{\rho_c}\biggr)^{q-1} \biggl(\frac{2q \Psi_0 y}{b^2} \biggr) \, . </math> </td> </tr> </table> This means that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\bold{u}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~ \frac{1}{\rho} \biggl\{ \boldsymbol{\hat\imath} \biggl[ \biggl( \frac{\rho}{\rho_c}\biggr)^{q-1} \biggl(\frac{2q \Psi_0 y}{b^2} \biggr) \biggr] - \boldsymbol{\hat\jmath} \biggl[ \biggl( \frac{\rho}{\rho_c}\biggr)^{q-1} \biggl(\frac{2q \Psi_0x}{a^2} \biggr) \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~\bold{u}\cdot \bold{u} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{\rho^2} \biggl\{ \biggl[ \biggl( \frac{\rho}{\rho_c}\biggr)^{q-1} \biggl(\frac{2q \Psi_0 y}{b^2} \biggr) \biggr]^2 + \biggl[ \biggl( \frac{\rho}{\rho_c}\biggr)^{q-1} \biggl(\frac{2q \Psi_0x}{a^2} \biggr) \biggr]^2 \biggr\} \, . </math> </td> </tr> </table> It also means that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\zeta_z </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{\partial }{\partial x}\biggl[\biggl( \frac{\rho}{\rho_c}\biggr)^{q-2} \biggl(\frac{2q \Psi_0 x}{\rho_c a^2} \biggr) \biggr] + \frac{\partial }{\partial y} \biggl[\biggl( \frac{\rho}{\rho_c}\biggr)^{q-2} \biggl(\frac{2q \Psi_0 y}{\rho_c b^2} \biggr) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl( \frac{\rho}{\rho_c}\biggr)^{q-2} \biggl(\frac{2q \Psi_0 }{\rho_c a^2} \biggr) ~+ ~ \biggl(\frac{2q \Psi_0 x}{\rho_c a^2} \biggr) \frac{\partial }{\partial x} \biggl( \frac{\rho}{\rho_c}\biggr)^{q-2} ~+ ~ \biggl( \frac{\rho}{\rho_c}\biggr)^{q-2} \biggl(\frac{2q \Psi_0 }{\rho_c b^2} \biggr) ~+ ~ \biggl(\frac{2q \Psi_0 y}{\rho_c b^2} \biggr) \frac{\partial }{\partial y} \biggl( \frac{\rho}{\rho_c}\biggr)^{q-2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl( \frac{\rho}{\rho_c}\biggr)^{q-2} \biggl(\frac{2q \Psi_0 }{\rho_c } \biggr) \biggl[ \frac{1}{a^2} + \frac{1}{b^2}\biggr] ~+ ~ \biggl(\frac{2q \Psi_0 x}{\rho_c a^2} \biggr) \biggl[ (q-2) \rho_c^{2-q} \rho^{q-3} \frac{\partial \rho}{\partial x}\biggr] ~+ ~ \biggl(\frac{2q \Psi_0 y}{\rho_c b^2} \biggr) \biggl[ (q-2) \rho_c^{2-q} \rho^{q-3} \frac{\partial \rho}{\partial y}\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl( \frac{\rho}{\rho_c}\biggr)^{q-2} \biggl(\frac{2q \Psi_0 }{\rho_c } \biggr) \biggl[ \frac{1}{a^2} + \frac{1}{b^2}\biggr] ~- ~ \biggl(\frac{2q \Psi_0 x}{\rho_c^2 a^2} \biggr) \biggl[ (q-2) \biggl(\frac{\rho}{\rho_c}\biggr)^{q-3} \frac{2\rho_c x}{a^2}\biggr] ~- ~ \biggl(\frac{2q \Psi_0 y}{\rho_c^2 b^2} \biggr) \biggl[ (q-2) \biggl(\frac{\rho}{\rho_c}\biggr)^{q-3} \frac{2\rho_c y}{b^2}\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{q\Psi_0}{\rho_c} \biggl\{ 2\biggl( \frac{\rho}{\rho_c}\biggr)^{q-2} \biggl[ \frac{1}{a^2} + \frac{1}{b^2}\biggr] ~- ~ \biggl(\frac{2 x}{ a^2} \biggr) \biggl[ (q-2) \biggl(\frac{\rho}{\rho_c}\biggr)^{q-3} \frac{2x}{a^2}\biggr] ~- ~ \biggl(\frac{2 y}{ b^2} \biggr) \biggl[ (q-2) \biggl(\frac{\rho}{\rho_c}\biggr)^{q-3} \frac{2 y}{b^2}\biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{q\Psi_0}{\rho_c} \biggl(\frac{\rho}{\rho_c}\biggr)^{q-3} \biggl[ 2\biggl( \frac{\rho}{\rho_c}\biggr) \biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) ~- ~ 4(q-2) \biggl(\frac{x^2}{ a^4} \biggr) ~- ~ 4(q-2) \biggl(\frac{y^2}{ b^4} \biggr) \biggr] </math> </td> </tr> </table> Hence, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~(\vec\zeta + 2\vec\Omega) \times \bold{u}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \boldsymbol{\hat\imath} \frac{\partial \Psi}{\partial x} + \boldsymbol{\hat\jmath} \frac{\partial \Psi}{\partial y} \biggr] \biggl\{ \frac{ 2\Omega_f}{\rho_c}\biggl( \frac{\rho}{\rho_c}\biggr)^{-1} + \biggl( \frac{\rho}{\rho_c}\biggr)^{-1} \frac{ \zeta_z }{\rho_c} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~ \biggl[ \boldsymbol{\hat\imath} \biggl( \frac{\rho}{\rho_c}\biggr)^{q-1} \biggl(\frac{2q \Psi_0 x}{a^2} \biggr) + \boldsymbol{\hat\jmath} \biggl( \frac{\rho}{\rho_c}\biggr)^{q-1} \biggl(\frac{2q \Psi_0 y}{b^2} \biggr) \biggr] \biggl\{ \frac{ 2\Omega_f}{\rho_c}\biggl( \frac{\rho}{\rho_c}\biggr)^{-1} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \frac{q\Psi_0}{\rho_c^2} \biggl(\frac{\rho}{\rho_c}\biggr)^{q-4} \biggl[ 2\biggl( \frac{\rho}{\rho_c}\biggr) \biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) ~- ~ 4(q-2) \biggl(\frac{x^2}{ a^4} \biggr) ~- ~ 4(q-2) \biggl(\frac{y^2}{ b^4} \biggr) \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~2q\Psi_0 \biggl[ \boldsymbol{\hat\imath} \biggl(\frac{x}{a^2} \biggr) + \boldsymbol{\hat\jmath} \biggl(\frac{y}{b^2} \biggr) \biggr] \biggl( \frac{\rho}{\rho_c}\biggr)^{q-2} \biggl\{ \frac{ 2\Omega_f}{\rho_c} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \frac{q\Psi_0}{\rho_c^2} \biggl(\frac{\rho}{\rho_c}\biggr)^{q-3} \biggl[ 2\biggl( \frac{\rho}{\rho_c}\biggr) \biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) ~- ~ 4(q-2) \biggl(\frac{x^2}{ a^4} \biggr) ~- ~ 4(q-2) \biggl(\frac{y^2}{ b^4} \biggr) \biggr] \biggr\} \, . </math> </td> </tr> </table> ====Exponent q = 2==== Notice that, if <math>~q = 2</math>, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[ (\vec\zeta + 2\vec\Omega) \times \bold{u} \biggr]_{q=2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~\frac{8\Psi_0}{\rho_c}\biggl\{ \Omega_f + \frac{2\Psi_0}{\rho_c} \biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) \biggr\} \biggl[ \boldsymbol{\hat\imath} \biggl(\frac{x}{a^2} \biggr) + \boldsymbol{\hat\jmath} \biggl(\frac{y}{b^2} \biggr) \biggr] \, . </math> </td> </tr> </table> Now, if we choose a function, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~F_B(\Psi)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{D_1}{2} \Psi^{1 / 2} = \frac{D_1}{2} \Psi_0^{1 / 2} \biggl(\frac{\rho}{\rho_c}\biggr) = \frac{D_1}{2} \Psi_0^{1 / 2} \biggl[1 - \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) \biggr] \, ,</math> </td> </tr> </table> we obtain, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\nabla F_B(\Psi)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~\frac{D_1}{2} \Psi_0^{1 / 2} \cdot \nabla \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~D_1 \Psi_0^{1 / 2} \biggl[ \boldsymbol{\hat\imath} \biggl(\frac{x}{a^2} \biggr) + \boldsymbol{\hat\jmath} \biggl(\frac{y}{b^2} \biggr) \biggr] \, . </math> </td> </tr> </table> This is consistent with the elliptic PDE constraint if, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~D_1 </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{8\Psi_0^{1 / 2}}{\rho_c}\biggl\{ \Omega_f + \frac{2\Psi_0}{\rho_c} \biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) \biggr\} \, .</math> </td> </tr> </table> Also if <math>~q = 2</math>, we have, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\bold{u}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{\rho} \biggl\{ \boldsymbol{\hat\imath} \frac{\partial \Psi}{\partial y} - \boldsymbol{\hat\jmath} \frac{\partial \Psi}{\partial x} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{\rho} \biggl\{ -\boldsymbol{\hat\imath} \biggl( \frac{\rho}{\rho_c}\biggr) \biggl(\frac{2q \Psi_0 y}{b^2} \biggr) + \boldsymbol{\hat\jmath} \biggl( \frac{\rho}{\rho_c}\biggr) \biggl(\frac{2q \Psi_0 x}{a^2} \biggr) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl\{ -\boldsymbol{\hat\imath} \biggl(\frac{4 \Psi_0 y}{\rho_c b^2} \biggr) + \boldsymbol{\hat\jmath} \biggl(\frac{4 \Psi_0 x}{\rho_c a^2} \biggr) \biggr\} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \bold{u}\cdot \bold{u}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl(\frac{4 \Psi_0 y}{\rho_c b^2} \biggr)^2 + \biggl(\frac{4 \Psi_0 x}{\rho_c a^2} \biggr)^2 </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl( \frac{4 \Psi_0}{\rho_c}\biggr)^2 \biggl[ \frac{x^2}{a^4} + \frac{y^2}{b^4} \biggr] \, . </math> </td> </tr> </table> Keep in mind that, [[#AndalibBernoulli|as discussed above]], we are trying to satisfy the scalar Bernoulli relation, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~C_B - H - \Phi_\mathrm{grav}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ F_B(\Psi) + \frac{1}{2}u^2 - \frac{1}{2}\Omega_f^2 (x^2 + y^2) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{4\Psi_0}{\rho_c}\biggl\{ \Omega_f + \frac{2\Psi_0}{\rho_c} \biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) \biggr\}\biggl[1 - \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) \biggr] + \frac{1}{2}\biggl( \frac{4 \Psi_0}{\rho_c}\biggr)^2 \biggl[ \frac{x^2}{a^4} + \frac{y^2}{b^4} \biggr] - \frac{1}{2}\Omega_f^2 (x^2 + y^2) \, . </math> </td> </tr> </table> The right-hand-side of this expression does not appear to be rich enough to balance the gravitational potential (on the left-hand-side) which, as [[#Trial_.232|detailed above]], contains <math>~x^2 y^2</math> and <math>~x^4</math> and <math>~y^4</math> terms. ====Exponent q = 3==== Alternatively, if <math>~q = 3</math>, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[ (\vec\zeta + 2\vec\Omega) \times \bold{u} \biggr]_{q=3}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~6\Psi_0 \biggl[ \boldsymbol{\hat\imath} \biggl(\frac{x}{a^2} \biggr) + \boldsymbol{\hat\jmath} \biggl(\frac{y}{b^2} \biggr) \biggr] \biggl( \frac{\rho}{\rho_c}\biggr) \biggl\{ \frac{ 2\Omega_f}{\rho_c} + \frac{3\Psi_0}{\rho_c^2} \biggl[ 2\biggl( \frac{\rho}{\rho_c}\biggr) \biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) ~- ~ 4 \biggl(\frac{x^2}{ a^4} \biggr) ~- ~ 4 \biggl(\frac{y^2}{ b^4} \biggr) \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~ 6\Psi_0 \biggl\{ \frac{ 2\Omega_f}{\rho_c} \biggl( \frac{\rho}{\rho_c}\biggr) - \frac{12\Psi_0}{\rho_c^2} \biggl[ \frac{x^2}{ a^4} + \frac{y^2}{ b^4} \biggr]\biggl( \frac{\rho}{\rho_c}\biggr) + \frac{6\Psi_0}{\rho_c^2} \biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) \biggl( \frac{\rho}{\rho_c}\biggr)^2 \biggr\} \boldsymbol{\hat{f}} \, , </math> </td> </tr> </table> where, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\boldsymbol{\hat{f}}</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~ \biggl[ \boldsymbol{\hat\imath} \biggl(\frac{x}{a^2} \biggr) + \boldsymbol{\hat\jmath} \biggl(\frac{y}{b^2} \biggr) \biggr] \, . </math> </td> </tr> </table> If we choose a function, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~F_B(\Psi)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ D_1\Psi^{2/3} + D_2 \Psi </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ D_1\Psi_0^{2/3} \biggl(\frac{\rho}{\rho_c}\biggr)^2 + D_2 \Psi_0 \biggl(\frac{\rho}{\rho_c}\biggr)^3 \, , </math> </td> </tr> </table> we obtain, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\nabla F_B(\Psi)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~\biggl[ 2D_1 \Psi_0^{2/3} \biggl(\frac{\rho}{\rho_c}\biggr) + 3D_2\Psi_0 \biggl( \frac{\rho}{\rho_c}\biggr)^2\biggr] \cdot \nabla \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~\biggl\{ 2D_1 \Psi_0^{2/3} \biggl[1 - \biggl(\frac{x^2}{a^2} + \frac{y^2}{b^2}\biggr) \biggr] + 3D_2\Psi_0 \biggl[1 - \biggl(\frac{x^2}{a^2} + \frac{y^2}{b^2}\biggr) \biggr]^2 \biggr\} \cdot \biggl[ \boldsymbol{\hat\imath} \biggl(\frac{2x}{a^2} \biggr) + \boldsymbol{\hat\jmath} \biggl(\frac{2y}{b^2} \biggr) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~\biggl\{ 2D_1 \Psi_0^{2/3} \biggl[1 - \biggl(\frac{x^2}{a^2} + \frac{y^2}{b^2}\biggr) \biggr] + 3D_2\Psi_0 \biggl[1 - 2\biggl(\frac{x^2}{a^2} + \frac{y^2}{b^2}\biggr) + \biggl(\frac{x^2}{a^2} + \frac{y^2}{b^2}\biggr)^2 \biggr] \biggr\} \cdot \biggl[ \boldsymbol{\hat\imath} \biggl(\frac{2x}{a^2} \biggr) + \boldsymbol{\hat\jmath} \biggl(\frac{2y}{b^2} \biggr) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~\biggl\{ \biggl(2D_1 \Psi_0^{2/3} + 3D_2\Psi_0 \biggr) - (2D_1 \Psi_0^{2/3} +6D_2\Psi_0)\biggl(\frac{x^2}{a^2} + \frac{y^2}{b^2}\biggr) + 3D_2\Psi_0 \biggl(\frac{x^2}{a^2} + \frac{y^2}{b^2}\biggr)^2 \biggr\} \boldsymbol{\hat{f}} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~\biggl\{ \biggl(2D_1 \Psi_0^{2/3} + 3D_2\Psi_0 \biggr) - (2D_1 \Psi_0^{2/3} +6D_2\Psi_0)\biggl(\frac{x^2}{a^2} + \frac{y^2}{b^2}\biggr) + 3D_2\Psi_0 \biggl[ \frac{x^4}{a^4} + 2\biggl( \frac{x^2 y^2}{a^2 b^2} \biggr) + \frac{y^4}{b^4}\biggr] \biggr\} \boldsymbol{\hat{f}} \, . </math> </td> </tr> </table> ---- Let's reorganize and expand the terms in both of these expressions in order to ascertain whether or not they can be matched. First … <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[ (\vec\zeta + 2\vec\Omega) \times \bold{u} \biggr]_{q=3}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~ 6\Psi_0 \biggl\{ \frac{ 2\Omega_f}{\rho_c} \biggl[ 1 - \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) \biggr] - \frac{12\Psi_0}{\rho_c^2} \biggl[ \frac{x^2}{ a^4} + \frac{y^2}{ b^4} \biggr] \biggl[ 1 - \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \frac{6\Psi_0}{\rho_c^2} \biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) \biggl[ 1 - \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) \biggr]^2 \biggr\} \boldsymbol{\hat{f}} \, , </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~ 6\Psi_0 \biggl\{ \frac{ 2\Omega_f}{\rho_c} - \frac{ 2\Omega_f}{\rho_c} \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) - \frac{12\Psi_0}{\rho_c^2} \biggl[ \frac{x^2}{ a^4} + \frac{y^2}{ b^4} \biggr] + \frac{12\Psi_0}{\rho_c^2} \biggl[ \frac{x^2}{ a^4} + \frac{y^2}{ b^4} \biggr] \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \frac{6\Psi_0}{\rho_c^2} \biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) \biggl[ 1 - 2\biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) + \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr)^2 \biggr] \biggr\} \boldsymbol{\hat{f}} \, , </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~ 6\Psi_0 \biggl\{ \biggl[ \frac{ 2\Omega_f}{\rho_c} +\frac{6\Psi_0}{\rho_c^2} \cdot \frac{(a^2 + b^2)}{a^2b^2} \biggr] - \biggl[ \frac{ 2\Omega_f}{\rho_c} + \frac{12\Psi_0}{\rho_c^2}\biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr)\biggr]\biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - \frac{12\Psi_0}{\rho_c^2} \biggl( \frac{x^2}{ a^4} + \frac{y^2}{ b^4} \biggr) + \frac{12\Psi_0}{\rho_c^2} \biggl( \frac{x^4}{ a^6} + \frac{x^2 y^2}{ a^4 b^2} + \frac{x^2 y^2}{ a^2 b^4} + \frac{y^4}{ b^6}\biggr) + \frac{6\Psi_0}{\rho_c^2}\biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr)\biggl( \frac{x^4}{a^4} + \frac{2x^2 y^2}{a^2 b^2}+ \frac{y^4}{b^4} \biggr) \biggr\} \boldsymbol{\hat{f}} \, , </math> </td> </tr> </table> In order for the zeroth-order terms to match, we need, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~2D_1 \Psi_0^{2/3} + 3D_2\Psi_0 </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 6\Psi_0 \biggl\{ \frac{ 2\Omega_f}{\rho_c} +\frac{6\Psi_0}{\rho_c^2} \cdot \frac{(a^2 + b^2)}{a^2b^2} \biggr\} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ 2D_1 \Psi_0^{2/3} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 3\Psi_0 \biggl\{ \frac{ 4\Omega_f \rho_c a^2 b^2}{\rho_c^2 a^2 b^2} +\frac{12\Psi_0(a^2 + b^2)}{\rho_c^2 a^2 b^2} \biggr\} - 3D_2\Psi_0 </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 3\Psi_0 \biggl\{ \frac{ 4\Omega_f \rho_c a^2 b^2}{\rho_c^2 a^2 b^2} +\frac{12\Psi_0(a^2 + b^2)}{\rho_c^2 a^2 b^2} -D_2\biggr\} \, . </math> </td> </tr> </table> Then we also need, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ 6D_2\Psi_0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~6\Psi_0 \biggl[ \frac{ 2\Omega_f}{\rho_c} + \frac{12\Psi_0}{\rho_c^2}\biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr)\biggr] - 2D_1 \Psi_0^{2/3} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~6\Psi_0 \biggl[ \frac{ 2\Omega_f}{\rho_c} + \frac{12\Psi_0}{\rho_c^2}\biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr)\biggr] - 3\Psi_0 \biggl\{ \frac{ 4\Omega_f \rho_c a^2 b^2}{\rho_c^2 a^2 b^2} +\frac{12\Psi_0(a^2 + b^2)}{\rho_c^2 a^2 b^2} -D_2\biggr\} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ 3D_2\Psi_0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~6\Psi_0 \biggl[ \frac{ 2\Omega_f \rho_c a^2b^2}{\rho_c^2 a^2 b^2} + \frac{12\Psi_0}{\rho_c^2}\frac{(a^2 + b^2)}{a^2b^2} \biggr] - 6\Psi_0 \biggl[ \frac{ 2\Omega_f \rho_c a^2 b^2}{\rho_c^2 a^2 b^2} +\frac{6\Psi_0(a^2 + b^2)}{\rho_c^2 a^2 b^2} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~6\Psi_0 \biggl[ \frac{6\Psi_0(a^2 + b^2)}{\rho_c^2a^2b^2}\biggr] </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ D_2</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{12\Psi_0(a^2 + b^2)}{\rho_c^2a^2b^2} \, . </math> </td> </tr> </table> Finally, we need, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~3D_2\Psi_0 \biggl[ \frac{x^4}{a^4} + 2\biggl( \frac{x^2 y^2}{a^2 b^2} \biggr) + \frac{y^4}{b^4}\biggr] \cdot \frac{1}{6\Psi_0} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - \frac{12\Psi_0}{\rho_c^2} \biggl( \frac{x^2}{ a^4} + \frac{y^2}{ b^4} \biggr) + \frac{12\Psi_0}{\rho_c^2} \biggl( \frac{x^4}{ a^6} + \frac{x^2 y^2}{ a^4 b^2} + \frac{x^2 y^2}{ a^2 b^4} + \frac{y^4}{ b^6}\biggr) + \frac{6\Psi_0}{\rho_c^2}\biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr)\biggl( \frac{x^4}{a^4} + \frac{2x^2 y^2}{a^2 b^2}+ \frac{y^4}{b^4} \biggr) </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \frac{(a^2 + b^2)}{a^2b^2} \biggl[ \frac{x^4}{a^4} + 2\biggl( \frac{x^2 y^2}{a^2 b^2} \biggr) + \frac{y^4}{b^4}\biggr] </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - 2 \biggl( \frac{x^2}{ a^4} + \frac{y^2}{ b^4} \biggr) + 2 \biggl( \frac{x^4}{ a^6} + \frac{x^2 y^2}{ a^4 b^2} + \frac{x^2 y^2}{ a^2 b^4} + \frac{y^4}{ b^6}\biggr) + \frac{(a^2 + b^2)}{a^2b^2} \biggl( \frac{x^4}{a^4} + \frac{2x^2 y^2}{a^2 b^2}+ \frac{y^4}{b^4} \biggr) </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~\biggl( \frac{x^2}{ a^4} + \frac{y^2}{ b^4} \biggr) </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl( \frac{x^4}{ a^6} + \frac{x^2 y^2}{ a^4 b^2} + \frac{x^2 y^2}{ a^2 b^4} + \frac{y^4}{ b^6}\biggr) </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~\frac{x^2}{ a^4} + \frac{y^2}{ b^4} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{x^4}{ a^6} + \frac{y^4}{ b^6} + \frac{x^2 y^2(a^2 + b^2) }{a^4 b^4} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ x^4 b^6 + y^4 a^6 </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ a^2 b^2 [x^2 b^4 - (a^2+b^2)x^2 y^2 + y^2a^4] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ a^2 b^2 [x^4 b^4 - (a^2+b^2)x^2 y^2 + y^4a^4] +a^2b^2[x^2b^4 + y^2a^4] -a^2b^2[x^4b^4 + y^4a^4] </math> </td> </tr> </table> ---- Keeping in mind that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Psi</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\Psi_0 \biggl( \frac{\rho}{\rho_c} \biggr)^q \, ,</math> </td> </tr> </table> and that, after setting <math>~q = 3</math>, we have chosen, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~F_B(\Psi)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ D_1\Psi^{2/3} + D_2 \Psi </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ D_1\Psi_0^{2/3} \biggl(\frac{\rho}{\rho_c}\biggr)^2 + D_2 \Psi_0 \biggl(\frac{\rho}{\rho_c}\biggr)^3 \, , </math> </td> </tr> </table> let's try again. <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[ (\vec\zeta + 2\vec\Omega) \times \bold{u} \biggr]_{q=3} - \nabla F_B(\Psi)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~ 6\Psi_0 \biggl\{ \frac{ 2\Omega_f}{\rho_c} \biggl( \frac{\rho}{\rho_c}\biggr) - \frac{12\Psi_0}{\rho_c^2} \biggl[ \frac{x^2}{ a^4} + \frac{y^2}{ b^4} \biggr]\biggl( \frac{\rho}{\rho_c}\biggr) + \frac{6\Psi_0}{\rho_c^2} \biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) \biggl( \frac{\rho}{\rho_c}\biggr)^2 \biggr\} \cdot \boldsymbol{\hat{f}} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ +~\biggl[ 2D_1 \Psi_0^{2/3} \biggl(\frac{\rho}{\rho_c}\biggr) + 3D_2\Psi_0 \biggl( \frac{\rho}{\rho_c}\biggr)^2\biggr] \cdot \nabla \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~ 3\Psi_0 \biggl\{ \frac{ 2\Omega_f}{\rho_c} - \frac{12\Psi_0}{\rho_c^2} \biggl[ \frac{x^2}{ a^4} + \frac{y^2}{ b^4} \biggr] + \frac{6\Psi_0}{\rho_c^2} \biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) \biggl( \frac{\rho}{\rho_c}\biggr) \biggr\} \cdot 2\biggl(\frac{\rho}{\rho_c}\biggr)\boldsymbol{\hat{f}} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ +~\biggl[ 2D_1 \Psi_0^{2/3} + 3D_2\Psi_0 \biggl( \frac{\rho}{\rho_c}\biggr)\biggr] \cdot 2\biggl(\frac{\rho}{\rho_c}\biggr)\boldsymbol{\hat{f}} \, . </math> </td> </tr> </table> Now, set … <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~2D_1 \Psi_0^{2/3}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{6\Psi_0 \Omega_f}{\rho_c}</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ D_1 </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{3\Psi_0^{1 / 3} \Omega_f}{\rho_c} \, ;</math> </td> </tr> </table> and, set … <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~3D_2 \Psi_0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{18 \Psi_0^2(a^2 + b^2)}{\rho_c^2 a^2 b^2} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ D_2 </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{6 \Psi_0(a^2 + b^2)}{\rho_c^2 a^2 b^2} \, . </math> </td> </tr> </table> We then have, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[ (\vec\zeta + 2\vec\Omega) \times \bold{u} \biggr]_{q=3} - \nabla F_B(\Psi)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~ 3\Psi_0 \biggl\{ - \frac{12\Psi_0}{\rho_c^2} \biggl[ \frac{x^2}{ a^4} + \frac{y^2}{ b^4} \biggr] \biggr\} \cdot 2\biggl(\frac{\rho}{\rho_c}\biggr)\boldsymbol{\hat{f}} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{72\Psi_0^2}{\rho_c^2} \biggl[ \frac{x^2}{ a^4} + \frac{y^2}{ b^4} \biggr] \cdot \biggl(\frac{\rho}{\rho_c}\biggr)\boldsymbol{\hat{f}} \, . </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ (\bold{u}\cdot \bold{u}) \nabla \biggl(\frac{\rho}{\rho_c}\biggr) \, . </math> </td> </tr> </table> <font color="red">'''VERY INTERESTING!''' (29 September 2020)</font> ====Exponent q = 4==== <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[ (\vec\zeta + 2\vec\Omega) \times \bold{u} \biggr]_{q=4}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~8\Psi_0 \biggl[ \boldsymbol{\hat\imath} \biggl(\frac{x}{a^2} \biggr) + \boldsymbol{\hat\jmath} \biggl(\frac{y}{b^2} \biggr) \biggr] \biggl( \frac{\rho}{\rho_c}\biggr)^{2} \biggl\{ \frac{ 2\Omega_f}{\rho_c} + \frac{8\Psi_0}{\rho_c^2} \biggl(\frac{\rho}{\rho_c}\biggr) \biggl[ \biggl( \frac{\rho}{\rho_c}\biggr) \biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) ~- ~ 4 \biggl(\frac{x^2}{ a^4} \biggr) ~- ~ 4 \biggl(\frac{y^2}{ b^4} \biggr) \biggr] \biggr\} \, . </math> </td> </tr> </table> ===Trial #4=== We begin with the, <div align="center"> Euler Equation<br /> written <font color="#770000">'''in terms of the Vorticity'''</font> and<br /> <font color="#770000">'''as viewed from a Rotating Reference Frame'''</font> <math>\frac{\partial \bold{u}}{\partial t} + (\boldsymbol\zeta+2{\vec\Omega}_f) \times {\bold{u}}= - \frac{1}{\rho} \nabla P - \nabla \biggl[\Phi + \frac{1}{2}u^2 - \frac{1}{2}|{\vec{\Omega}}_f \times \vec{x}|^2 \biggr]</math> . </div> Next, we rewrite this expression to incorporate the following three realizations: <ul> <li>For a barotropic fluid, the term involving the pressure gradient can be replaced with a term involving the enthalpy via the relation, <math>~\nabla H = \nabla P/\rho</math>.</li> <li>The expression for the centrifugal potential can be rewritten as, <math>~\tfrac{1}{2}|\vec\Omega_f \times \vec{x}|^2 = \tfrac{1}{2}\Omega_f^2 (x^2 + y^2)</math>.</li> <li>In steady state, <math>~\partial \bold{u}/\partial t = 0</math>.</li> </ul> This means that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ (\boldsymbol\zeta+2{\vec\Omega}_f) \times {\bold{u}}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>- \nabla \biggl[H + \Phi_\mathrm{grav} + \frac{1}{2}u^2 - \frac{1}{2}\Omega_f^2 (x^2 + y^2) \biggr] \, .</math> </td> </tr> </table> If the term on the left-hand-side of this equation can be expressed in terms of the gradient of a scalar function, then it can be readily grouped with all the other terms on the right-hand-side, which already are in the gradient form. Building on the insight that we have gained from the [[#Exponent_q_.3D_3|above examination of systems for which the exponent, q = 3]], let's change the <math>~\tfrac{1}{2}\nabla u^2</math> term on the RHS to <math>~\tfrac{1}{2}\nabla (\rho u)^2</math> then reexamine the LHS. <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ [ (\boldsymbol\zeta+2{\vec\Omega}_f) \times {\bold{u}} ]_{q=3}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>- \nabla \biggl[H + \Phi_\mathrm{grav} - \frac{1}{2}\Omega_f^2 (x^2 + y^2) \biggr] - \nabla \biggl[\frac{1}{2}u^2\biggr]</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> - \nabla \biggl[H + \Phi_\mathrm{grav} - \frac{1}{2}\Omega_f^2 (x^2 + y^2) \biggr] - \biggl(\frac{\rho}{\rho_c}\biggr)^{-2} \biggl\{ \biggl(\frac{\rho}{\rho_c}\biggr)^{2} \nabla \biggl[\frac{1}{2}u^2\biggr]\biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> - \nabla \biggl[H + \Phi_\mathrm{grav} - \frac{1}{2}\Omega_f^2 (x^2 + y^2) \biggr] - \biggl(\frac{\rho}{\rho_c}\biggr)^{-2} \biggl\{ \nabla \biggl[\frac{1}{2}\biggl(\frac{\rho}{\rho_c}\biggr)^{2} u^2\biggr] - \frac{1}{2}u^2\nabla \biggl[\biggl(\frac{\rho}{\rho_c}\biggr)^{2}\biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ [ (\boldsymbol\zeta+2{\vec\Omega}_f) \times {\bold{u}} ]_{q=3} - \nabla F_B</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> - \nabla \biggl[H + \Phi_\mathrm{grav} + F_B - \frac{1}{2}\Omega_f^2 (x^2 + y^2) \biggr] - \biggl(\frac{\rho}{\rho_c}\biggr)^{-2} \biggl\{ \nabla \biggl[\frac{1}{2}\biggl(\frac{\rho}{\rho_c}\biggr)^{2} u^2\biggr] - \frac{1}{2}u^2\nabla \biggl[\biggl(\frac{\rho}{\rho_c}\biggr)^{2}\biggr] \biggr\} \, . </math> </td> </tr> </table> Now, rewriting the LHS gives, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~[ (\boldsymbol\zeta+2{\vec\Omega}_f) \times {\bold{u}} ]_{q=3} - \nabla F_B</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~ 3\Psi_0 \biggl\{ \frac{ 2\Omega_f}{\rho_c} - \frac{12\Psi_0}{\rho_c^2} \biggl[ \frac{x^2}{ a^4} + \frac{y^2}{ b^4} \biggr] + \frac{6\Psi_0}{\rho_c^2} \biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) \biggl( \frac{\rho}{\rho_c}\biggr) \biggr\} \cdot 2\biggl( \frac{\rho}{\rho_c}\biggr)\boldsymbol{\hat{f}} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ +~\biggl\{ 2D_1 \Psi_0^{2/3} + 3D_2\Psi_0 \biggl( \frac{\rho}{\rho_c}\biggr) \biggr\} \cdot 2\biggl( \frac{\rho}{\rho_c}\biggr)\boldsymbol{\hat{f}} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~3\Psi_0 \biggl\{ \frac{12\Psi_0}{\rho_c^2} \biggl[ \frac{x^2}{ a^4} + \frac{y^2}{ b^4} \biggr] \biggr\} \cdot 2\biggl( \frac{\rho}{\rho_c}\biggr)\boldsymbol{\hat{f}} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~ \frac{36\Psi_0^2}{\rho_c^2} \biggl[ \frac{x^2}{ a^4} + \frac{y^2}{ b^4} \biggr] \cdot \nabla \biggl(\frac{\rho}{\rho_c}\biggr)^2 \, , </math> </td> </tr> </table> where we have set, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~D_1 = \frac{3\Psi_0^{1 / 3} \Omega_f}{\rho_c}</math> </td> <td align="center"> and, </td> <td align="left"> <math>~D_2 = \frac{6\Psi_0 (a^2 + b^2)}{\rho_c^2 a^2 b^2} \, .</math> </td> </tr> </table> Notice that when the exponent, <math>~q=3</math>, we have, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~[ \bold{u}\cdot \bold{u} ]_{q=3} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{\rho^2} \biggl\{ \biggl[ \biggl( \frac{\rho}{\rho_c}\biggr)^{q-1} \biggl(\frac{2q \Psi_0 y}{b^2} \biggr) \biggr]^2 + \biggl[ \biggl( \frac{\rho}{\rho_c}\biggr)^{q-1} \biggl(\frac{2q \Psi_0x}{a^2} \biggr) \biggr]^2 \biggr\}_{q=3} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{\rho^2} \biggl\{ \biggl[ \biggl( \frac{\rho}{\rho_c}\biggr)^{2} \biggl(\frac{6 \Psi_0 y}{b^2} \biggr) \biggr]^2 + \biggl[ \biggl( \frac{\rho}{\rho_c}\biggr)^{2} \biggl(\frac{6 \Psi_0x}{a^2} \biggr) \biggr]^2 \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{36\Psi_0^2}{\rho_c^2} \biggl( \frac{\rho}{\rho_c}\biggr)^2 \biggl[ \frac{x^2}{a^4} + \frac{ y^2}{b^4} \biggr] \, . </math> </td> </tr> </table> Hence, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~[ (\boldsymbol\zeta+2{\vec\Omega}_f) \times {\bold{u}} ]_{q=3} - \nabla F_B</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~ \biggl( \frac{\rho}{\rho_c}\biggr)^{-2} u^2 \cdot \nabla \biggl(\frac{\rho}{\rho_c}\biggr)^2 \, . </math> </td> </tr> </table> ===Trial#5=== Let's return to the above-mentioned, <div align="center"> <font color="#770000">'''Eulerian Representation'''</font><br /> of the Euler Equation <br /> <font color="#770000">'''as viewed from a Rotating Reference Frame'''</font> <math>\frac{\partial \bold{u}}{\partial t} + (\bold{u}\cdot \nabla) \bold{u} = - \frac{1}{\rho} \nabla P - \nabla \Phi_\mathrm{grav} - {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x}) - 2{\vec{\Omega}}_f \times \bold{u} \, .</math> </div> In steady state, this can be rewritten as, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~(\bold{u}\cdot \nabla) \bold{u} + 2{\vec{\Omega}}_f \times \bold{u}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - \nabla \biggl[H + \Phi_\mathrm{grav} - \frac{1}{2}|{\vec{\Omega}}_f \times \vec{x}|^2 \biggr] \, . </math> </td> </tr> </table> Let's focus on the left-hand-side, which is expressed entirely in terms of the rotating-frame velocity, <math>~\bold{u}</math>, and the (constant) angular frequency of rotation of the coordinate frame, <math>~\Omega_f</math>. Rewriting the LHS, we have, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> '''LHS:''' <math>~2{\vec{\Omega}}_f \times \bold{u} + (\bold{u}\cdot \nabla) \bold{u}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2\Omega_f \biggl[ \boldsymbol{\hat\jmath} u_x - \boldsymbol{\hat\imath} u_y\biggr] + \biggl[ u_x \frac{\partial}{\partial x} + u_y \frac{\partial}{\partial y} \biggr] \biggl[\boldsymbol{\hat\imath} u_x + \boldsymbol{\hat\jmath} u_y \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2\Omega_f \biggl[ \boldsymbol{\hat\jmath} u_x - \boldsymbol{\hat\imath} u_y\biggr] + \boldsymbol{\hat\imath}\biggl[ u_x \frac{\partial u_x}{\partial x} + u_y \frac{\partial u_x}{\partial y} \biggr] + \boldsymbol{\hat\jmath} \biggl[ u_x \frac{\partial u_y}{\partial x} + u_y \frac{\partial u_y}{\partial y} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2\Omega_f \biggl[ \boldsymbol{\hat\jmath} u_x - \boldsymbol{\hat\imath} u_y\biggr] + \biggl[\boldsymbol{\hat\imath} u_x \frac{\partial u_x}{\partial x} + \boldsymbol{\hat\jmath} u_y \frac{\partial u_y}{\partial y} \biggr] + \biggl[ \boldsymbol{\hat\imath}u_y \frac{\partial u_x}{\partial y} + \boldsymbol{\hat\jmath} u_x \frac{\partial u_y}{\partial x} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2\Omega_f \biggl[ \boldsymbol{\hat\jmath} u_x - \boldsymbol{\hat\imath} u_y\biggr] + \frac{1}{2}\biggl\{ \biggl[\boldsymbol{\hat\imath} \frac{\partial u_x^2}{\partial x} + \boldsymbol{\hat\jmath} \frac{\partial u_y^2}{\partial y} \biggr] + \biggl[\boldsymbol{\hat\imath} \frac{\partial u_y^2}{\partial x} + \boldsymbol{\hat\jmath} \frac{\partial u_x^2}{\partial y} \biggr] \biggr\} + \biggl[ \boldsymbol{\hat\imath}u_y \frac{\partial u_x}{\partial y} + \boldsymbol{\hat\jmath} u_x \frac{\partial u_y}{\partial x} \biggr] - \frac{1}{2}\biggl[\boldsymbol{\hat\imath} \frac{\partial u_y^2}{\partial x} + \boldsymbol{\hat\jmath} \frac{\partial u_x^2}{\partial y} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2\Omega_f \biggl[ \boldsymbol{\hat\jmath} u_x - \boldsymbol{\hat\imath} u_y\biggr] + \frac{1}{2}\nabla u^2 + \boldsymbol{\hat\imath} u_y \biggl[ \frac{\partial u_x}{\partial y} - \frac{\partial u_y}{\partial x} \biggr] + \boldsymbol{\hat\jmath} u_x \biggl[ \frac{\partial u_y}{\partial x} - \frac{\partial u_x}{\partial y} \biggr] \, . </math> </td> </tr> </table> Next, to the extent possible, let's express the LHS in terms of the dimensionless mass density, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\sigma \equiv \frac{\rho}{\rho_c}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~1 - \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2}\biggr) \, .</math> </td> </tr> </table> <table border="1" align="center" cellpadding="10" width="80%"><tr><td align="left"> We will assume that the stream-function, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Psi</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\Psi_0 \sigma^q \, ,</math> </td> </tr> </table> in which case, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\rho_c \sigma \bold{u}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\boldsymbol{\hat\imath} \frac{\partial\Psi}{\partial y} - \boldsymbol{\hat\jmath} \frac{\partial\Psi}{\partial x}</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ \rho_c \bold{u}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{q}{(q-1)} \biggl[ \boldsymbol{\hat\imath} \frac{\partial\sigma^{q-1}}{\partial y} - \boldsymbol{\hat\jmath} \frac{\partial\sigma^{q-1}}{\partial x} \biggr] \, .</math> </td> </tr> </table> That is, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\rho_c u_x</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{q}{(q-1)} \biggl( \frac{\partial\sigma^{q-1}}{\partial y} \biggr) </math> </td> <td align="center"> and, </td> <td align="right"> <math>~\rho_c u_y</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- \frac{q}{(q-1)} \biggl( \frac{\partial\sigma^{q-1}}{\partial x} \biggr) \, .</math> </td> </tr> </table> </td></tr></table> The first term on the LHS becomes, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ 2\Omega_f \biggl[ \boldsymbol{\hat\jmath} u_x - \boldsymbol{\hat\imath} u_y\biggr] </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{2\Omega_f}{\rho_c} \cdot \frac{q}{q-1} \biggl[ \boldsymbol{\hat\jmath} \biggl( \frac{\partial\sigma^{q-1}}{\partial y} \biggr) + \boldsymbol{\hat\imath} \biggl( \frac{\partial\sigma^{q-1}}{\partial x} \biggr)\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{2\Omega_f}{\rho_c} \cdot \frac{q}{q-1} \nabla (\sigma^{q-1}) \, . </math> </td> </tr> </table> The third term on the LHS becomes, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ \boldsymbol{\hat\imath} u_y \biggl[ \frac{\partial u_x}{\partial y} - \frac{\partial u_y}{\partial x} \biggr] </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \boldsymbol{\hat\imath} \biggl( \frac{u_y}{\rho_c} \biggr) \biggl\{ \frac{\partial }{\partial y} \biggl[ \frac{q}{(q-1)} \biggl( \frac{\partial\sigma^{q-1}}{\partial y} \biggr) \biggr] + \frac{\partial }{\partial x} \biggl[ \frac{q}{(q-1)} \biggl( \frac{\partial\sigma^{q-1}}{\partial x} \biggr) \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{q}{(q-1)} \cdot \boldsymbol{\hat\imath} \biggl( \frac{u_y}{\rho_c} \biggr) \biggl\{ \nabla^2 \sigma^{q-1} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~\biggl[ \frac{q}{\rho_c(q-1)}\biggr]^2 \biggl\{ \nabla^2 \sigma^{q-1} \cdot \boldsymbol{\hat\imath} \biggl( \frac{\partial\sigma^{q-1}}{\partial x} \biggr) \biggr\} </math> </td> </tr> </table> Similarly, the fourth term on the LHS becomes, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ \boldsymbol{\hat\jmath} u_x \biggl[ \frac{\partial u_y}{\partial x} - \frac{\partial u_x}{\partial y} \biggr] </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~\boldsymbol{\hat\jmath} \biggl( \frac{u_x}{\rho_c} \biggr) \biggl\{ \frac{\partial }{\partial x} \biggl[ \frac{q}{(q-1)} \biggl( \frac{\partial\sigma^{q-1}}{\partial x} \biggr) \biggr] + \frac{\partial }{\partial y} \biggl[ \frac{q}{(q-1)} \biggl( \frac{\partial\sigma^{q-1}}{\partial y} \biggr) \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~ \frac{q}{(q-1)} \cdot \boldsymbol{\hat\jmath} \biggl( \frac{u_x}{\rho_c} \biggr) \biggl\{ \nabla^2 \sigma^{q-1} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~ \biggl[ \frac{q}{\rho_c(q-1)}\biggr]^2\biggl\{ \nabla^2 \sigma^{q-1} \cdot \boldsymbol{\hat\jmath} \biggl( \frac{\partial \sigma^{q-1}}{\partial y} \biggr) \biggr\} \, . </math> </td> </tr> </table> <table border="1" width="80%" cellpadding="10" align="center"><tr><td align="left"> Note that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\partial \sigma^{q-1}}{\partial x_i}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ (q-1)\sigma^{q-2} \frac{\partial \sigma}{\partial x_i} = - (q-1)\sigma^{q-2} \biggl( \frac{2x_i}{a_i^2} \biggr) </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ (\rho_c u)^2 = \rho_c^2(u_x^2 + u_y^2)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{q^2}{(q-1)^2}\biggl\{ \biggl[ - (q-1)\sigma^{q-2} \biggl( \frac{2x}{a^2} \biggr) \biggr]^2 + \biggl[ - (q-1)\sigma^{q-2} \biggl( \frac{2y}{b^2} \biggr) \biggr]^2 \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 4q^2 \sigma^{2(q-2)} \biggl[\frac{x^2}{a^4} +\frac{y^2}{b^4} \biggr] \, . </math> </td> </tr> </table> </td></tr></table> Hence, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\nabla^2 \sigma^{q-1}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{\partial}{\partial x} \biggl[ \frac{\partial \sigma^{q-1}}{\partial x}\biggr] + \frac{\partial}{\partial y} \biggl[ \frac{\partial \sigma^{q-1}}{\partial y}\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - (q-1)\biggl\{ \frac{\partial}{\partial x} \biggl[ \sigma^{q-2} \biggl( \frac{2x}{a^2} \biggr)\biggr] + \frac{\partial}{\partial y} \biggl[ \sigma^{q-2} \biggl( \frac{2y}{b^2} \biggr)\biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - (q-1)\biggl\{ \sigma^{q-2} \frac{\partial}{\partial x} \biggl[ \biggl( \frac{2x}{a^2} \biggr)\biggr] + \biggl( \frac{2x}{a^2} \biggr)\frac{\partial}{\partial x} \biggl[ \sigma^{q-2} \biggr] + \sigma^{q-2}\frac{\partial}{\partial y} \biggl[ \biggl( \frac{2y}{b^2} \biggr)\biggr] + \biggl( \frac{2y}{b^2} \biggr)\frac{\partial}{\partial y} \biggl[ \sigma^{q-2} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - (q-1)\biggl\{ \biggl[ \frac{2}{a^2} + \frac{2}{b^2} \biggr] \sigma^{q-2} + \frac{2x(q-2)}{a^2} \biggl[ \sigma^{q-3} \frac{\partial \sigma}{\partial x}\biggr] + \frac{2y(q-2)}{b^2} \biggl[ \sigma^{q-3} \frac{\partial \sigma}{\partial y}\biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - (q-1) \biggl\{ \biggl[ \frac{2}{a^2} + \frac{2}{b^2} \biggr] \sigma^{q-2} - 2 (q-2)\sigma^{q-3} \biggl[ \frac{2x^2}{a^4} +\frac{2y^2}{b^4} \biggr] \biggr\} \, . </math> </td> </tr> </table> So, when they are added together, the third and fourth terms give, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ \boldsymbol{\hat\imath} u_y \biggl[ \frac{\partial u_x}{\partial y} - \frac{\partial u_y}{\partial x} \biggr] + \boldsymbol{\hat\jmath} u_x \biggl[ \frac{\partial u_y}{\partial x} - \frac{\partial u_x}{\partial y} \biggr] </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~\biggl[ \frac{q}{\rho_c(q-1)}\biggr]^2 \nabla^2 \sigma^{q-1} \biggl\{ \boldsymbol{\hat\imath} \biggl( \frac{\partial\sigma^{q-1}}{\partial x} \biggr) +~ \boldsymbol{\hat\jmath} \biggl( \frac{\partial \sigma^{q-1}}{\partial y} \biggr) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \frac{2q^2}{\rho_c^2(q-1)}\biggr] \biggl\{ \biggl[ \frac{1}{a^2} + \frac{1}{b^2} \biggr] \sigma^{q-2} - 2 (q-2)\sigma^{q-3} \biggl[ \frac{x^2}{a^4} +\frac{y^2}{b^4} \biggr] \biggr\} \biggl\{ \boldsymbol{\hat\imath} \biggl( \frac{\partial\sigma^{q-1}}{\partial x} \biggr) +~ \boldsymbol{\hat\jmath} \biggl( \frac{\partial \sigma^{q-1}}{\partial y} \biggr) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \frac{2q^2}{\rho_c^2(q-1)}\biggr]\biggl[ \frac{1}{a^2} + \frac{1}{b^2} \biggr] \biggl\{ \sigma^{q-2} \biggr\} \biggl\{ \boldsymbol{\hat\imath} \biggl( \frac{\partial\sigma^{q-1}}{\partial x} \biggr) +~ \boldsymbol{\hat\jmath} \biggl( \frac{\partial \sigma^{q-1}}{\partial y} \biggr) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -~\biggl[ \frac{4(q-2)q^2}{\rho_c^2(q-1)}\biggr] \biggl\{ \biggl[\frac{(\rho_c u)^2}{4q^2}\biggr] \sigma^{1-q} \biggr\}\biggl\{ \boldsymbol{\hat\imath} \biggl( \frac{\partial\sigma^{q-1}}{\partial x} \biggr) +~ \boldsymbol{\hat\jmath} \biggl( \frac{\partial \sigma^{q-1}}{\partial y} \biggr) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{2q^2}{\rho_c^2(2q-3)} \biggl[ \frac{1}{a^2} + \frac{1}{b^2} \biggr] \nabla \sigma^{2q-3} -~\frac{(q-2)(\rho_c u)^2}{\rho_c^2} \nabla\ln(\sigma) \, . </math> </td> </tr> </table> Hence, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> '''LHS:''' <math>~2{\vec{\Omega}}_f \times \bold{u} + (\bold{u}\cdot \nabla) \bold{u}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{2q\Omega_f}{\rho_c(q-1)} \nabla (\sigma^{q-1}) + \frac{1}{2}\nabla u^2 + \frac{2q^2}{\rho_c^2(2q-3)} \biggl[ \frac{1}{a^2} + \frac{1}{b^2} \biggr] \nabla \sigma^{2q-3} -~(q-2)u^2 \nabla\ln(\sigma) \, . </math> </td> </tr> </table> <table border="1" width="80%" cellpadding="10" align="center"><tr><td align="left"> Note that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ \frac{1}{2}\nabla u^2 -~(q-2)u^2 \nabla\ln(\sigma) </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{2} \biggl\{ \nabla u^2 -~u^2 \nabla\ln[\sigma^{2(q-2)}] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{u^2}{2} \biggl\{ \nabla \ln [u^2] -~ \nabla\ln[\sigma^{2(q-2)}] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{u^2}{2} \biggl\{ \nabla \ln \biggl[ \frac{u^2}{ \sigma^{2(q-2)} } \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{u^2}{2} \biggl\{ \nabla \ln \biggl[ \biggl( \frac{2q}{ \rho_c } \biggr)^2 \biggl( \frac{x^2}{a^4} + \frac{y^2}{b^4} \biggr)\biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{2q^2}{\rho_c^2} \sigma^{2(q-2)} \biggl[\frac{x^2}{a^4} +\frac{y^2}{b^4} \biggr] \biggl\{ \nabla \ln \biggl[ \biggl( \frac{2q}{ \rho_c } \biggr)^2 \biggl( \frac{x^2}{a^4} + \frac{y^2}{b^4} \biggr)\biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{\sigma^{2(q-2)}}{2} \biggl\{ \nabla \biggl[ \biggl( \frac{2q}{ \rho_c } \biggr)^2 \biggl( \frac{x^2}{a^4} + \frac{y^2}{b^4} \biggr)\biggr] \biggr\} \, . </math> </td> </tr> </table> Or, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ \frac{1}{2}\nabla u^2 +~(2-q)u^2 \nabla\ln(\sigma) </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{2} \biggl\{ \nabla u^2 +~u^2 \nabla\ln[\sigma^{2(2-q)}] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl( \frac{1}{2}\biggr)\sigma^{-2(2-q)} \biggl\{ \sigma^{2(2-q)}\nabla u^2 +~u^2 \nabla[\sigma^{2(2-q)}] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl( \frac{1}{2}\biggr)\sigma^{2(q-2)} \biggl\{ \nabla[u^2 \sigma^{2(2-q)}] \biggr\} \, . </math> </td> </tr> </table> </td></tr></table> ====Exponent q = 2==== <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> '''LHS:''' <math>~[2{\vec{\Omega}}_f \times \bold{u} + (\bold{u}\cdot \nabla) \bold{u}]_{q=2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{4\Omega_f}{\rho_c} \nabla \sigma + \frac{1}{2}\nabla u^2 + \frac{8}{\rho_c^2} \biggl[ \frac{1}{a^2} + \frac{1}{b^2} \biggr] \nabla \sigma \, . </math> </td> </tr> </table> ====Exponent q = 3==== <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> '''LHS:''' <math>~[2{\vec{\Omega}}_f \times \bold{u} + (\bold{u}\cdot \nabla) \bold{u} ]_{q=3}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{3\Omega_f}{\rho_c} \nabla \sigma^{2} + \frac{1}{2}\nabla u^2 + \frac{6}{\rho_c^2} \biggl[ \frac{1}{a^2} + \frac{1}{b^2} \biggr] \nabla \sigma^{3} -~u^2 \nabla\ln(\sigma) \, . </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{3\Omega_f}{\rho_c} \nabla \sigma^{2} + \frac{6}{\rho_c^2} \biggl[ \frac{1}{a^2} + \frac{1}{b^2} \biggr] \nabla \sigma^{3} + u^2\nabla \ln\biggl(\frac{u}{\sigma}\biggr) \, . </math> </td> </tr> </table> ===Trial #6=== <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> '''LHS:''' <math>~2{\vec{\Omega}}_f \times \bold{u} + (\bold{u}\cdot \nabla) \bold{u}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2\Omega_f \biggl[ \boldsymbol{\hat\jmath} u_x - \boldsymbol{\hat\imath} u_y\biggr] + \frac{1}{2}\nabla u^2 + \boldsymbol{\hat\imath} u_y \biggl[ \frac{\partial u_x}{\partial y} - \frac{\partial u_y}{\partial x} \biggr] + \boldsymbol{\hat\jmath} u_x \biggl[ \frac{\partial u_y}{\partial x} - \frac{\partial u_x}{\partial y} \biggr] \, . </math> </td> </tr> </table> <table border="1" align="center" cellpadding="10" width="80%"><tr><td align="left"> We will assume that the stream-function, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Psi</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\Psi_0 (\alpha - \sigma )^q </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \frac{\partial\Psi}{\partial x_i}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~\Psi_0 q(\alpha - \sigma )^{q-1} \frac{\partial\sigma}{\partial x_i} \, ,</math> </td> </tr> </table> in which case, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\rho_c \sigma \bold{u}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\boldsymbol{\hat\imath} \frac{\partial\Psi}{\partial y} - \boldsymbol{\hat\jmath} \frac{\partial\Psi}{\partial x}</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~\boldsymbol{\hat\imath} \Psi_0 q(\alpha - \sigma )^{q-1} \frac{\partial\sigma}{\partial y} + \boldsymbol{\hat\jmath}~\Psi_0 q(\alpha - \sigma )^{q-1} \frac{\partial\sigma}{\partial x} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ \rho_c \bold{u}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~\boldsymbol{\hat\imath} \biggl[ \frac{\Psi_0 q(\alpha - \sigma )^{q-1} }{\sigma} \biggr]\frac{\partial\sigma}{\partial y} + \boldsymbol{\hat\jmath}~\biggl[ \frac{\Psi_0 q(\alpha - \sigma )^{q-1}}{\sigma}\biggr] \frac{\partial\sigma}{\partial x} \, .</math> </td> </tr> </table> That is, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\rho_c u_x</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~\biggl[ \frac{\Psi_0 q(\alpha - \sigma )^{q-1} }{\sigma} \biggr]\frac{2y}{b^2} </math> </td> <td align="center"> and, </td> <td align="right"> <math>~\rho_c u_y</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \frac{\Psi_0 q(\alpha - \sigma )^{q-1}}{\sigma}\biggr] \frac{2x}{a^2} \, .</math> </td> </tr> </table> </td></tr></table> The first term on the LHS becomes, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ 2\Omega_f \biggl[ \boldsymbol{\hat\jmath} u_x - \boldsymbol{\hat\imath} u_y\biggr] </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{2\Omega_f}{\rho_c} \biggl[- \frac{\Psi_0 q(\alpha - \sigma )^{q-1}}{\sigma}\biggr] \biggl[\boldsymbol{\hat\jmath} \biggl( \frac{2y}{b^2} \biggr) + \boldsymbol{\hat\imath} \biggl( \frac{2x}{a^2} \biggr)\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ + \frac{2\Omega_f}{\rho_c} \biggl[ \frac{\Psi_0 q(\alpha - \sigma )^{q-1}}{\sigma}\biggr] \nabla \sigma \, . </math> </td> </tr> </table> ===Trial #7=== ====Uncluttered Setup==== Let's simply look at the vortensity expression as defined in [[#Part_II|Part II, above]], namely, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~g(\Psi)</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\frac{(2\Omega_f + \zeta_z)}{\rho_c\sigma} \, ,</math> </td> </tr> </table> and recognize that we are ultimately interested in the function, <math>~F_B(\Psi)</math>, defined such that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{dF_B(\Psi)}{d\Psi}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-g(\Psi) \, .</math> </td> </tr> </table> We start with the expression for the z-component of the vorticity, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\zeta_z </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~\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> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~\biggl\{ \frac{\partial }{\partial x}\biggl[ \biggl( \frac{\Psi_0}{\rho_c \sigma} \biggr) \frac{\partial \sigma^q}{\partial x} \biggr] + \frac{\partial }{\partial y} \biggl[ \biggl( \frac{\Psi_0}{\rho_c\sigma} \biggr) \frac{\partial \sigma^q}{\partial y} \biggr] \biggr\} \, . </math> </td> </tr> </table> Next, appreciate that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{1}{\sigma} \frac{\partial \sigma^q}{\partial x_i}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ q \sigma^{q-2} \frac{\partial \sigma}{\partial x_i} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl( \frac{q}{q-1} \biggr) \frac{\partial \sigma^{q-1}}{\partial x_i} \, . </math> </td> </tr> </table> Hence, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\zeta_z </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~ \frac{q\Psi_0}{\rho_c (q-1) } \biggl\{ \frac{\partial }{\partial x}\biggl[ \frac{\partial \sigma^{q-1}}{\partial x} \biggr] + \frac{\partial }{\partial y} \biggl[ \frac{\partial \sigma^{q-1}}{\partial y} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~ \biggl[ \frac{q\Psi_0}{\rho_c (q-1) } \biggr] \nabla^2\sigma^{q-1} \, . </math> </td> </tr> </table> So, the vortensity is, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~g(\sigma) = \frac{(2\Omega_f + \zeta_z )}{\rho_c\sigma}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl(\frac{2\Omega_f}{\rho_c}\biggr)\sigma^{-1} - \biggl[ \frac{q\Psi_0}{\rho_c^2 (q-1) } \biggr] \sigma^{-1} \nabla^2\sigma^{q-1} \, . </math> </td> </tr> </table> Let's switch to the stream-function via the ''assumed'' relation, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\sigma</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl( \frac{\Psi}{\Psi_0}\biggr)^{1/q} \, ;</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ g(\Psi) </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl(\frac{2\Omega_f}{\rho_c}\biggr)\biggl( \frac{\Psi}{\Psi_0}\biggr)^{-1/q} - \biggl[ \frac{q\Psi_0}{\rho_c^2 (q-1) } \biggr] \biggl( \frac{\Psi}{\Psi_0}\biggr)^{-1/q} \nabla^2\biggl( \frac{\Psi}{\Psi_0}\biggr)^{(q-1)/q} \, . </math> </td> </tr> </table> This expression gives the vortensity in what appears to be the desired form — that is, expressed strictly in terms of the stream function, <math>~\Psi</math> — for a wide range of values of the exponent, <math>~q</math>. [<font color="red">'''CAUTION:'''</font> the <math>~\nabla^2</math> operator is an exception.] It is not yet (13 October 2020) clear to me how — or if — the second term on the right-hand-side of this expression can be integrated to give <math>~F(\Psi)</math>. But the first term can be obtained from, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~F_{1^\mathrm{st}}(\Psi)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -\biggl(\frac{2\Omega_f}{\rho_c}\biggr) \frac{q}{(q-1)} \biggl( \frac{\Psi}{\Psi_0}\biggr)^{(q-1)/q} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \frac{dF_{1^\mathrm{st}}(\Psi)}{d\Psi}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -\biggl(\frac{2\Omega_f}{\rho_c}\biggr) \biggl( \frac{\Psi}{\Psi_0}\biggr)^{-1/q} \, . </math> </td> </tr> </table> ====T5 Coordinates==== Let's evaluate the <math>~\nabla^2</math> operator by expressing it and its argument in terms of [[Appendix/Ramblings/EllipticCylinderCoordinates#T5_Coordinates|T5 Coordinates]]. Note that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\lambda_1</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~(x^2 + \kappa^2 y^2)^{1 / 2} \, ,</math> </td> </tr> </table> where, <math>~\kappa \equiv a/b</math>. The specified density distribution is, therefore, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\sigma \equiv \frac{\rho}{\rho_c}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~1 - \frac{\lambda_1^2}{a^2} \, ,</math> </td> </tr> </table> and the stream-function is, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\Psi}{\Psi_0}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\sigma^q = \biggl[ 1 - \frac{\lambda_1^2}{a^2} \biggr]^q \, .</math> </td> </tr> </table> The relevant [[Appendix/Ramblings/EllipticCylinderCoordinates#Laplacian|T5-Coordinate System Laplacian]] is, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\nabla^2 \mathfrak{F}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \frac{1}{\lambda_1^2 \ell^2} \biggr] \biggl[ \frac{\partial^2 \mathfrak{F}}{\partial \lambda_1^2}\biggr] - \biggl[ \frac{1}{\lambda_1^3 \ell^2} \biggr] \frac{\partial \mathfrak{F}}{\partial \lambda_1} + \biggl[ \frac{(1 + \kappa^2)}{\lambda_1 } \biggr] \frac{\partial \mathfrak{F}}{\partial \lambda_1} \, , </math> </td> </tr> </table> where, <div align="center"> <math>~\ell \equiv (x^2 + \kappa^4 y^2)^{- 1 / 2} \, ,</math> </div> and in the present context, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathfrak{F}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl(\frac{\Psi}{\Psi_0}\biggr)^{(q-1)/q} = \biggl[ 1 - \frac{\lambda_1^2}{a^2} \biggr]^{q-1} \, . </math> </td> </tr> </table> Hence, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\partial \mathfrak{F}}{\partial \lambda_1}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - \frac{2(q-1)\lambda_1}{a^2} \biggl[ 1 - \frac{\lambda_1^2}{a^2} \biggr]^{q-2} = - \biggl[ \frac{2(q-1)\lambda_1}{a^2}\biggr]\biggl( \frac{\Psi}{\Psi_0} \biggr)^{(q-2)/q} \, ; </math> </td> </tr> <tr> <td align="right"> <math>~\frac{\partial^2 \mathfrak{F}}{\partial \lambda_1^2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - \frac{2(q-1)}{a^2} \biggl[ 1 - \frac{\lambda_1^2}{a^2} \biggr]^{q-2} + \frac{2(q-1)\lambda_1}{a^2} \biggl[ 1 - \frac{\lambda_1^2}{a^2} \biggr]^{q-3} \biggl[ \frac{2(q-2) \lambda_1 }{ a^2 } \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - \frac{2(q-1)}{a^2} \biggl( \frac{\Psi}{\Psi_0} \biggr)^{(q-2)/q} + \frac{4(q-1)(q-2)\lambda_1^2}{a^4} \biggl( \frac{\Psi}{\Psi_0} \biggr)^{(q-3)/q} \, . </math> </td> </tr> </table> And, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\nabla^2 \mathfrak{F}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \frac{1}{\lambda_1^2 \ell^2} \biggr] \biggl\{ - \frac{2(q-1)}{a^2} \biggl( \frac{\Psi}{\Psi_0} \biggr)^{(q-2)/q} + \frac{4(q-1)(q-2)\lambda_1^2}{a^4} \biggl( \frac{\Psi}{\Psi_0} \biggr)^{(q-3)/q} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - \biggl[ \frac{1}{\lambda_1^3 \ell^2} \biggr] \biggl\{ - \biggl[ \frac{2(q-1)\lambda_1}{a^2}\biggr]\biggl( \frac{\Psi}{\Psi_0} \biggr)^{(q-2)/q} \biggr\} + \biggl[ \frac{(1 + \kappa^2)}{\lambda_1 } \biggr] \biggl\{ - \biggl[ \frac{2(q-1)\lambda_1}{a^2}\biggr]\biggl( \frac{\Psi}{\Psi_0} \biggr)^{(q-2)/q} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{4(q-1)(q-2)}{a^4\ell^2} \biggl( \frac{\Psi}{\Psi_0} \biggr)^{(q-3)/q} - \biggl[ \frac{2(q-1)(1 + \kappa^2)}{a^2}\biggr]\biggl( \frac{\Psi}{\Psi_0} \biggr)^{(q-2)/q} \, . </math> </td> </tr> </table> ====All Together==== Putting this all together, we obtain, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ g(\Psi) </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl(\frac{2\Omega_f}{\rho_c}\biggr)\biggl( \frac{\Psi}{\Psi_0}\biggr)^{-1/q} - \biggl[ \frac{q\Psi_0}{\rho_c^2 (q-1) } \biggr] \biggl( \frac{\Psi}{\Psi_0}\biggr)^{-1/q} \biggl\{ \frac{4(q-1)(q-2)}{a^4\ell^2} \biggl( \frac{\Psi}{\Psi_0} \biggr)^{(q-3)/q} - \biggl[ \frac{2(q-1)(1 + \kappa^2)}{a^2}\biggr]\biggl( \frac{\Psi}{\Psi_0} \biggr)^{(q-2)/q} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl(\frac{2\Omega_f}{\rho_c}\biggr)\biggl( \frac{\Psi}{\Psi_0}\biggr)^{-1/q} + \biggl[ \frac{2q(1 + \kappa^2)\Psi_0}{\rho_c^2 a^2} \biggr] \biggl( \frac{\Psi}{\Psi_0} \biggr)^{(q-3)/q} - \biggl[ \frac{4 q(q-2) \Psi_0}{\rho_c^2 a^4} \biggr] (x^2 + \kappa^4 y^2) \biggl( \frac{\Psi}{\Psi_0} \biggr)^{(q-4)/q} \, . </math> </td> </tr> </table> Note that, for the [[Appendix/Ramblings/EllipticCylinderCoordinates#Invert_Coordinate_Mapping|specific ''example'' case of]] <math>~\kappa^2 = 2</math>, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[ \frac{1}{\ell^2} \biggr]_{\kappa^2=2} = (x^2 + 4y^2)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ x^2 \Lambda = \frac{\Lambda \lambda_2^2}{2} \biggl[ \Lambda - 1 \biggr] = 2\lambda_1^2 \biggl[ \frac{\Lambda}{\Lambda + 1} \biggr] \, . </math> </td> </tr> </table> where, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Lambda</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\biggl[ 1 + \frac{4\lambda_1^2}{\lambda_2^2} \biggr]^{1 / 2} \, .</math> </td> </tr> </table>
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