rigid-rigid boundaries

We translate the origin to the mid height of the chamber and the problem to be solved is now given by

$$(D^2-a^2)^3W=-a^2 R_a W$$ (56)

subject to boundary condition

$$W=DW = (D^2-a^2)^2 W = 0 \quad z=0,1$$ (57)

The problem is symmetric with respect to the two boundaries so the eigen functions fall into two distinct classes: (even mode) one with vertical velocity symmetry with respect to the mid plane and the (odd mode) other with vertical velocity antisymmetry. Even mode has one row of cells along the vertical while odd has two row of cells. Let us assume the solution of (56) of the form

$$W = e^{qz}$$

where the roots $q$ are given by

$$(q^2-a^2)^3 = - R a^2$$ (58)

Let $R a^2= \lambda^3 a^6$ then the roots of (58) are given by

$$q^2 = -a^2(\lambda-1) \quad q^2=a^2[1+\frac{1}{2}\lambda(1\pm i\frac{\sqrt{3}}{2})]$$ (59)

Taking the square roots again, the roots are given by

$$\pm iq_0,\quad \pm q \quad \pm q^*$$ (60)

where

$$q_0 = a\sqrt{\lambda-1}$$

and

$$(q)=q_1=a\left[\frac{1}{2}\sqrt{1+\lambda+\lambda^2}+\frac{1}{2}(1+\frac{1}{2}\lambda)^{1/2}\right]^{1/2}$$

$$(q)=q_2=a\left[\frac{1}{2}\sqrt{1+\lambda+\lambda^2}-\frac{1}{2}(1+\frac{1}{2}\lambda)^{1/2}\right]^{1/2}$$

Here $q^*$ denote the complex conjugate of $q$. From these we have the following relations

$$(q_0^2+a^2)^2=a^4\lambda^2$$

$$(q^2-a^2)^2 = \frac{1}{2}a^4\lambda^2(-1\pm i\sqrt{3})$$

Even solution
The even solution is given by

$$W = A \cos(q_0 z) + B \cosh(q z) + C \cosh(q* z)$$ (61)

From (61) we get

$$DW = -A q_0\sin(q_0 z) + B q\sinh(q z) + C q^*\sinh(q* z)$$

$$(D^2-a^2)^2W = A (q_0^2+a^2)^2\cos(q_0 z) + B (q^2-a^2)^2\cosh(q z) + C (q^{*2}-a^2)^2\cosh(q* z)$$

The boundary conditions (57) then gives

$$\left[\begin{array}{ccc} \cos(q_0/2) &\cosh(q/2) &\cosh(q^*/2) \\... .../2) \end{array} \right] \left[\begin{array}{c} A\\ B\\ C \end{array}\right] = 0$$ (62)

For the nontrivial solution (after some manipulations) we must have

$$\left\vert\begin{array}{ccc} 1 &1 &1\\ -q_0\tan(q_0/2) & q\tanh(q... ...frac{1}{2}(i\sqrt{3}-1) & -\frac{1}{2}(i\sqrt{3}+1) \end{array} \right\vert = 0$$ (63)

which on simplification gives

$$\left[(\sqrt{3}+i)q\tanh(q/2)\right]+q_0 \tan(q_0/2) = 0$$ (64)

which can be written as (by further simplification)

$$-q_0\tan(q_0/2) = \frac{(q_1+q_2\sqrt{3})\sinh(q_1)+(q_1\sqrt{3}-q_2)\sin(q_2)}{\cosh(q_1)+\cos(q_2)}$$

This equation has to be solved by trial and error method: for a given value of $a$ we need to find the value of $\lambda$ and then we find the value of $R_a$. The value of $a_c$ and $R_c$ (Reid & Harris, Phys of Fluids, Vol-1) are given by

$$a_c=3.117 \quad R_c=1707.762$$ (65)

Taking $A_0=1$ in $C=B^*$ we can find $W$ and $\Theta$.

Odd solution
The odd solution is given by

$$W = A \sin(q_0 z) + B \sinh(q z) + C \sinh(q* z)$$ (66)

Proceeding as before the we obtain

$$q_0\cot(q_0/2) = \frac{(q_1+q_2\sqrt{3})\sinh(q_1)-(q_1\sqrt{3}-q_2)\sin(q_2)}{\cosh(q_1)-\cos(q_2)}$$

In this case minimum Rayleigh number occurs at $a=5.365$ and the corresponding value of the Rayleigh number is $R=17610.39$.