ASTR/ATOC-5410: Fluid Instabilities, Waves, and Turbulence August 31, 2016, Axel Brandenburg

Handout 4: Rayleigh–Bénard problem (Part II)

Most of the effort in comparing with laboratory measurements went into the treatment of suitable boundary conditions. Let us consider here the no-slip condition, i.e.,

u_{x} = u_{y} = u_{z} = 0  (1)

Owing to ∇ · u = 0, this implies u_{z,z} = 0, in addition to u_{z} = 0. Such a function can no longer be represented by simple sine and cosine series. Let us discuss here consequences for the stability analysis.

1 Normal mode analysis

One usually speaks of normal mode analysis, when the eigenfunction is decomposed into a complete set of functions. For the time being, we continue using a Fourier decomposition, but now only in the horizontal direction, so we set u_{1z} = û_{1z}(z) e^{σt+ik_{⊥}·x_{⊥}}, Let us inset this into Eq. (15) from Handout 3, i.e.,

(\frac{∂}{∂t} − ∇^{2}) (Pr\frac{∂}{∂t} − ∇^{2}) ∇^{2}u_{z1} = Ra ∇_{⊥}^{2}u_{z1}.  (2)

This yields

(σ + k_{⊥}^{2} − D^{2}) (Pr σ + k_{⊥}^{2} − D^{2}) (k_{⊥}^{2} − D^{2}) û_{1z}(z) = Ra k_{⊥}^{2}û_{1z}(z),  (3)

where D = ∂/∂z has been introduced as a shorthand; this is not to be confused with the advective derivative used earlier.

Another trick that can be invoked is what is called the principle of the exchange of stabilities, which really just means that σ is real and that the marginal states are characterized by σ = 0. We discussed this in Handout 3, but didn’t talk about exchange of stabilities. Chandrasekhar (1961) talks a lot about it and gives in his Section 11 a general proof of this for Rayleigh-Bénard convection in the absence of rotation. In the presence of rotation, however, the principle of the exchange of stabilities is not valid.

Thus, putting σ = 0 in Equation (3), and multiplying by −1 (so the coefficient in front of the highest derivative is positive) we have

(D^{2} − k_{⊥}^{2})^{3} û_{1z}(z) = −Ra k_{⊥}^{2}û_{1z}(z).  (4)

Note that this equation, which describes only the onset of convection, is independent of Pr. We have seen this before where the marginal stability condition for stress-free boundary conditions was independent of Pr.

The general solution can now we written as a superposition of solutions of the form

û_{1z}(z) = ∑_{±i=1}^{3} A_{i}e^{q_{i}z}  (5)

with, in general, complex values of q_{i}. Inserting this into Equation (4) yields

(q_{i}^{2} − k_{⊥}^{2})^{3} = −Ra k_{⊥}^{2}  (6)

for i = 1, 2, and 3. To solve this equation, we need to find the three roots of this equation. The footnote^{1}

e^{+iπ/3} = \frac{1}{2} + \frac{i}{2} \sqrt{3},  (7)
e^{−iπ/3} = \frac{1}{2} − \frac{i}{2} \sqrt{3},  (8)

1To find the three roots of (−1)^{1/3}, it is useful to represent −1 in the form −1 = e^{iπ}. The three solutions are then and

e^{iπ} = −1.  (9)

Likewise, if we wanted to find the roots of (−1)^{1/5}, for example, they would be given by e^{±iπ/5} = cos π/5 ± sin π/5, e^{±3iπ/5} = cos 3π/5 ± sin 3π/5, and, again, e^{5iπ/5} = −1.

is a reminder of how you do this. With these preparations, we can now write

q_{i}^{2} − k_{⊥}^{2} = Ra^{1/3} k_{⊥}^{2/3} × \{ \begin{matrix}−1 & for i = 1 \\ \frac{1}{2} + \frac{i}{2} \sqrt{3} & for i = 2 \\ \frac{1}{2} − \frac{i}{2} \sqrt{3} & for i = 3 \end{matrix}  (10)

for the three roots of q_{i}^{2}. To find all six roots of q_{i}, we begin with the simplest case, i.e.,

q_{±1} = ±\sqrt{k_{⊥}^{2} − Ra^{1/3} k_{⊥}^{2/3}} = ±i\sqrt{Ra^{1/3} k_{⊥}^{2/3} − k_{⊥}^{2}} = ±ik_{⊥}\sqrt{(Ra/k_{⊥}^{4})^{1/3} − 1}.  (11)

Next, we have

q_{±2} = ±\sqrt{k_{⊥}^{2} + Ra^{1/3} k_{⊥}^{2/3} (\frac{1}{2} + \frac{i}{2} \sqrt{3})} = ±k_{⊥}\sqrt{1 + \frac{1}{2} (Ra/k_{⊥}^{4})^{1/3} (1 + i\sqrt{3})}  (12)

and finally, q_{±3} is just given by the complex conjugate of q_{±2}, i.e.,

q_{±3} = q_{±2}^{∗}.  (13)

To construct the final solution and to determine the critical excitation condition, we need to invoke boundary conditions. In addition to those discussed in the preamble, i.e., û_{1z} = Dû_{1z} = 0, we still have the condition T̂ = 0, which can be expressed in terms of û_{1z} using Eq. (10) of Handout 3, which reduces to

(D^{2} − k_{⊥}^{2})^{2} û_{1z}(z) = 0  (14)

For each of the three pairs, the functions can be readily combined into a function that is symmetric around 0 by

û_{1z}(z) = ∑_{±i=1}^{3} A_{i}e^{q_{i}z} = ∑_{i=1}^{3} A_{i} (e^{q_{i}z} + e^{−q_{i}z}) = 2 ∑_{i=1}^{3} A_{i} cosh q_{i}z.  (15)

To obey the boundary condition û_{1z}(±1/2) = 0, we have to require that

∑_{i=1}^{3} cosh q_{i}/2 = 0.  (16)

This is one equation for the three unknowns A_{i} for i = 1, 2, and 3. Next, to obey the boundary condition Dû_{1z}(±1/2) = 0, we have to require that

∑_{i=1}^{3} sinh q_{i}/2 = 0.  (17)

Finally, to obey the boundary condition (D^{2} − k_{⊥}^{2})^{2}û_{1z}(±1/2) = 0, we have to require that

∑_{i=1}^{3} (q_{i}^{2} − k_{⊥}^{2})^{2} cosh q_{i}/2 = 0.  (18)

We now have 3 equations for the three unknowns A_{i} for i = 1, 2, and 3. This leads to a 3 × 3 matrix equation, where the eigenvector is given by (A_{1}, A_{2}, A_{3}) and the matrix is a function of Ra and k_{⊥}^{2}. The determinant of this matrix must vanish, which then results in a function Ra = Ra(k_{⊥}^{2}); see Fig. 11.10 of KCD. The smallest value of Ra is reached at k_{⊥} = 3.12 and gives Ra(k_{⊥}) = 1708.

References

Chandrasekhar, S. Hydrodynamic and Hydromagnetic Stability. Dover Publications, New York (1961).

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