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The subject material of this question is relevant to Rayleigh-Benard convection and the Boussinesq equations. It also appears relevant to deriving what people call an amplitude equation.

I am trying to go through separating $$(\nu \nabla^2\circ \nabla^2 - s \nabla^2)\eta(x,y,z)=0 \tag{1}$$ by using the separation of variables approach, $\nu$ and $s$ are constants.

So, I am trying to find equations for the possible $f$ and $g$ in $$(\nu \nabla^2\circ \nabla^2 - s \nabla^2)f(x,y)g(z)=0 \tag{2}$$

It looks to me, now , as if the separation cannot be performed but based upon the Spiegel lecture I mention, it looked as if the separated equations should be

\begin{align} \nabla_1^2f(x,y)&=-k^2 \\ [s-\nu(D^2-k^2)]\circ(D^2-k^2)g(z)&=0 \end{align} where \begin{align}\nabla_1^2&= \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\\ D^2&= \frac{\partial^2}{\partial z^2} \end{align} BUT I am having trouble working this through. Might anyone help with this?

OTHER INFORMATION

See 'Dissipative Boussinesq Dynamics' by Edward A Spiegel, at

http://www.dtic.mil/docs/citations/ADA114366

In particular, see Equation (6), pg. 34 of the PDF.

According to the above lecture, starting from the Boussinesq approximation and applying what has been termed a linearization it is possible to derive an equation for a function representing the vertical vorticity $\zeta(x,y,z,t)$ which is given as $$(\partial_t - \nu \nabla^2)\circ \nabla^2\zeta(x,y,z,t)=0$$

Looking for solutions of the form $$\zeta\propto e^{st}g_s(x,y,z)$$ and substitution of this into the partial differential equation, we have

\begin{align} (\partial_t - \nu \nabla^2)\circ \nabla^2e^{st}g_s(x,y,z)&=0\\ ( \partial_t e^{st}\nabla^2 - \nu e^{st} \nabla^2 \circ \nabla^2 )g_s&=0\\ ( \partial_t e^{st}\nabla^2 g_s - \nu e^{st} \nabla^2 \circ \nabla^2 g_s )&=0\\ \left( \frac{ \partial_t e^{st}\nabla^2 g_s } { \nu e^{st} \nabla^2 g_s } - \frac{ \nu e^{st} \nabla^2 \circ \nabla^2 g_s }{ \nu e^{st} \nabla^2 g_s } \right)&=0\\ \left( \frac{ \partial_t e^{st} } { \nu e^{st} } - \frac{ \nabla^2 \circ \nabla^2 g_s }{ \nabla^2 g_s } \right)&=0\\ \left( \frac{ \partial_t e^{st} } { e^{st} } - \frac{ \nu \nabla^2 \circ \nabla^2 g_s }{ \nabla^2 g_s } \right)&=0\\ \frac{ \partial_t e^{st} } { e^{st} } &= \frac{ \nu \nabla^2 \circ \nabla^2 g_s }{ \nabla^2 g_s } =C \end{align}

hence \begin{align} \partial_t e^{st} &= C e^{st} \\ \nu \nabla^2 \circ \nabla^2 g_s & =C \nabla^2 g_s\\ \end{align}

From the above, the separation constant $C$ must equal $s$, so our seprarated equations may be written

\begin{align} \partial_t e^{st} &= s e^{st} \\ (\nu \nabla^2 \circ \nabla^2 -s \nabla^2 ) g_s (x,y,z) & = 0\\ \end{align}

Putting $g_s(x,y,z)=\eta(x,y,z)$, in the above, gives (1).

To separate $\eta(x,y,z)$ such that $\eta(x,y,z)=f(x,y)g(z)$ we need to reexpress $$ (\nu \nabla^2 \circ \nabla^2 -s \nabla^2 ) \eta(x,y,z) = 0 $$ as

( terms involving only x and y ) + ( terms involving only z ) = 0

Put \begin{align} \nabla_1^2&= \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\\ D^2&= \frac{\partial^2}{\partial z^2}\\ \text{so }\nabla^2&=\nabla_1^2+D^2\\ \end{align}

Now trying to find consistent equations for the possible $f$ and $g$ in $$(\nu \nabla^2\circ \nabla^2 - s \nabla^2)f(x,y)g(z)=0 $$ we must have

\begin{align} (\nu (\nabla_1^2+D^2) \circ (\nabla_1^2+D^2) - s (\nabla_1^2+D^2) )f(x,y)g(z)&=0 \\ ( \nabla_1^2 \circ \nabla_1^2 + 2 \nabla_1^2 \circ D^2 +D^2 \circ D^2 ) fg - \frac{s} {\nu} \nabla_1^2fg - \frac{s} {\nu}D^2 fg&=0 \\ \end{align}

Using the following, \begin{align} \nabla_1^2f(x,y)&=-k^2 \\ \text {we have ; } ( -2 k^2 D^2 g +D^2 \circ D^2 fg) + k^2\frac{s} {\nu} g - \frac{s} {\nu}D^2 fg&=0 \\ ( - \frac{2 k^2}{f} D^2 g +D^2 \circ D^2 g) +\frac{ k^2}{f} \frac{s} {\nu} g - \frac{s} {\nu}D^2 g&=0 \\ \end{align}

This is as far as I get, perhaps there is an error in Spiegel's lecture and the separation cannot be performed? Perhaps $f$ can be approximated as equal to one or put equal to zero, for the purposes of the lecture material?

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Using the following condition, should lead to a separation into two equations.

$$ \nabla_1^2f(x,y)=-k^2 f(x,y) $$

It looks as if the separated equations can be written as \begin{align} \nabla_1^2f(x,y)&=-k^2 f(x,y) \\ [s-\nu(D^2-k^2)](D^2-k^2)g(z)&=0 \end{align} where \begin{align} \nabla_1^2&= \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\\ D^2&= \frac{\partial^2}{\partial z^2} \end{align}

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