
Rayleigh-Bénard
Storyboard 
One of the widely studied instabilities is the so-called Rayleigh Bernard instability that describes the structure of cells that forms when a liquid enters convection in a temperature gradient.
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Rayleigh–Bénard instability
Equation 
When a liquid in a gravitational field with acceleration g is exposed to two horizontal plates with upper temperature T_t and lower temperature T_b, where the lower temperature is greater than the upper temperature, a spontaneous flow is generated if the Rayleigh number
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exceeds a critical value. In this case, L represents the distance between the plates, d is the diffusivity, \alpha is the thermal expansion coefficient, and $
u$ is the kinematic viscosity.
You can find more information at this link: http://home.iitk.ac.in/~sghorai/NOTES/benard/benard.html
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Rayleigh-Bénard instability, limit free-free edges
Equation 
The eigenvalue problem becomes
(D^2-a^2)^3W=-a^2R_aW
subject to the boundary condition
$W=D^2W=D^4W=0 at
From this, we can show that
D^{2m}W=0
for
z=0,1
and
m = 1,2,\cdots
It follows that the required solution must be
W=A\sin(n\pi z)
with
n=1,2,3,\cdots
where A is a constant and n is an integer. Substituting W leads to the eigenvalue relationship
R_a=\frac{(n^2\pi^2+a^2)^3}{a^2}
For a given
a^2
, the lowest value of
R_c
occurs when
n=1
which is the lowest mode:
R_a=\frac{(\pi^2+a^2)^3}{a^2}
The critical Rayleigh number
R_c
is obtained by finding the minimum value of
R_a
as
a^2
varies.
\frac{dR_a}{da^2}=0\rightarrow a_c=\frac{\pi}{\sqrt{2}}
and the corresponding
R_c
is given by
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Rayleigh-Bénard instability, fixed-fixed edge limit
Equation 
If we set the origin at the midpoint of the chamber, the problem to be solved is given by
(D^2-a^2)^3W=-a^2R_aW
subject to the boundary condition
W=DW=(D^2-a^2)^2W=0
at
z=0,1
.
The problem is symmetric with respect to the two boundaries, so the eigenfunctions are divided into two distinct classes: (even mode), which has vertical velocity symmetry with respect to the midplane, and (odd mode), which has vertical velocity antisymmetry. The even mode has a single row of cells along the vertical, while the odd mode has two rows of cells. Let\'s assume the solution is of the form
W=e^{qz}
where the roots q are given by
(q^2-a^2)^3=-R_a^2
Let
R_a^2=\lambda^3a^6
then the roots are given by
q^2=-a^2(\lambda-1)
and
q^2=a^2\left[1+\frac{1}{2}\frac{\sqrt{3}}{2}\right]
Taking the square root again, the roots are \pm iq_0, \pm q, and \pm q^*, where q_0=a\sqrt{\lambda-1} and
re(q)=q_1=a\left[\frac{1}{2}\sqrt{1+\lambda+\lambda^2}+\frac{1}{2}\sqrt{1+\frac{\lambda}{2}}\right]^{1/2}
im(q)=q_2=a\left[\frac{1}{2}\sqrt{1+\lambda+\lambda^2}-\frac{1}{2}\sqrt{1+\frac{\lambda}{2}}\right]^{1/2}
Here, q^* denotes the complex conjugate of q. From these relationships, we have
(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 mode solution
The even mode solution is expressed as
W=A\cos(q_0z)+B\cosh(qz)+C\cosh(q^*z)
Therefore, we have
DW=-Aq_0\sin(q_0z)+Bq\sinh(qz)+Cq^*\sinh(q^*z)
(D^2-a^2)^2W=A(q_0^2+a^2)^2\cos(q_0z)+B(q^2-a^2)+C(q^{*2}-a^2)^2\cosh(q^*z)
The boundary conditions provide
\left[\begin{array}{ccc}\cos(q_0/2) & \cosh(q/2) & \cosh(q^/2) \ \sin(q_0/2) & \sinh(q/2) & \sinh(q^/2) \ -q_0\tan(q_0/2) & q\tanh(q/2) & q^\tanh(q^/2)\end{array}\right]\left[\begin{array}{c}A \ B \ C\end{array}\right]=0
For a nontrivial solution (after some manipulations), we must have
\displaystyle\left\vert\begin{array}{ccc} 1 & 1 & 1 \ -q_0\tan(q_0/2) & q\tanh(q/2) & q^\tanh(q^/2) \ \frac{1}{2}(i\sqrt{3}+1) & \frac{1}{2}(i\sqrt{3}-1) & -1\end{array}\right\vert=0
which, in simplified terms,
im\left[(\sqrt{3}+i)q\tanh(q/2)\right]+q_0\tan(q_0/2)=0
which can be written as (with 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 must be solved using the trial and error method: for a given value of a, we need to find the value of \lambda and then find the value of R_a. The critical values of a and R_a (Reid & Harris, Phys of Fluids, Vol-1) are given by
$a_c=3.117 and
Taking A_0=1 and C=B^*, we can find W and \Theta.
Impar Solution
The odd solution is represented as
W=A\sin(q_0z)+B\sinh(qz)+C\sinh(q^*z)
Proceeding similarly, we obtain
q_0\cot(q_0/2)=\displaystyle\frac{(q_1+q_2\sqrt{3})\sinh(q_1)-(q_1\sqrt{3}-q_2)\sin(q_2)}{\cosh(q_1)-\cosh(q_2)}
In this case, the minimum Rayleigh number occurs at a=5.365, and the corresponding value of the Rayleigh number is
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.
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Rayleigh-Bénard instability, fixed-free edge limit
Equation 
The solution for the case where the upper surface is free and the lower surface is rigid can be deduced from the odd solution of the rigid-rigid box.
The problem is defined by:
\frac{(D^2-a^2)^3}{W}=-a^2R_aW
subject to the boundary condition:
W = DW = (D^2-a^2)^2W = 0
at
z = 0
W = D^2W = D^4W = 0
at
z = 1
The boundary conditions at the mid-height for the odd solution hold. Therefore, an odd solution for the rigid-rigid limit at depth
d
provides a solution for the rigid-free limit at depth
d/2
. Thus, using the stability results from the rigid-rigid case, we have:
a_c = 5.365 / 2 \approx 2.682
and
R_c = 17610.39 / 2^4 \approx 1100.65
[Note: From the dimensional form of
\Delta_1^2f + k^2f
, we have
a = kL
(where
L
is the scale length) as the dimensionless wavenumber.]
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