# Peeter Joot's (OLD) Blog.

• ## Archives

 Adam C Scott on avoiding gdb signal noise… Ken on Scotiabank iTrade RESP …… Alan Ball on Oops. Fixing a drill hole in P… Peeter Joot's B… on Stokes theorem in Geometric… Exploring Stokes The… on Stokes theorem in Geometric…

• 290,315

## PHY454H1S Continuum Mechanics. Lecture 21: Stability. Taught by Prof. K. Das.

Posted by peeterjoot on March 30, 2012

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

# Stability. Some graphical illustrations.

What do we mean by stability? A configuration is stable if after a small disturbance it returns to it’s original position. A couple systems to consider are shown in figures (\ref{fig:continuumL21:continuumL21Fig1a}), (\ref{fig:continuumL21:continuumL21Fig1b}) and (\ref{fig:continuumL21:continuumL21Fig1c}).

Figure 1: Stable well configuration.

Figure 2: Instable peak configuration.

Figure 3: Stable tabletop configuration.

We can examine how a displacement $\delta x$ changes with time after making it. In a stable configuration without friction we will induce an oscillation as plotted in figure (\ref{fig:continuumL21:continuumL21Fig2a}) for the parabolic configuration. With friction we’ll have a damping effect. This is plotted for the parabolic well in figure (\ref{fig:continuumL21:continuumL21Fig2b}).

Figure 4: Displacement time evolution in undamped well system.

Figure 5: Displacement time evolution in damped well system.

For the inverted parabola our displacement takes the form of figure (\ref{fig:continuumL21:continuumL21Fig3})

Figure 6: Time evolution of displacement in instable parabolic configuration.

For the ball on the table, assuming some friction that stops the ball, fairly quickly, we’ll have a displacement as illustrated in figure (\ref{fig:continuumL21:continuumL21Fig3b})

Figure 7: Time evolution of displacement in tabletop configuration.

# Characterizing stability

Let’s suppose that our displacement can be described in exponential form

\begin{aligned}\delta x \sim e^{\sigma t}\end{aligned} \hspace{\stretch{1}}(3.1)

where $\sigma$ is the \textit{growth rate of perturbation}, and is in general a complex number of the form

\begin{aligned}\sigma = \sigma_\text{R} + i \sigma_\text{I}\end{aligned} \hspace{\stretch{1}}(3.2)

## Case I. Oscillatory unstability

A system of the form

\begin{aligned}\sigma_{\text{R}} &= 0 \\ \sigma_{\text{I}} &> 0\end{aligned}

\textit{oscillatory unstable}. An example of this is the undamped parabolic system illustrated above.

## Case II. Marginal unstability.

\begin{aligned}\delta x &\sim e^{\sigma_{\text{R}} t} e^{i \sigma_{\text{I}} t} \\ &\sim e^{\sigma_{\text{R}} t} \left( \cos \sigma_{\text{I}} t + i \sin \sigma_{\text{I}} t \right)\end{aligned}

We’ll call systems of the form

\begin{aligned}\sigma_{\text{I}} &= 0 \\ \sigma_{\text{R}} &> 0\end{aligned}

\textit{marginally unstable}. We can have unstable systems with $\sigma_{\text{I}} \ne 0$ but still $\sigma_{\text{R}} > 0$, but these are less common.

## Case III. Neutral stability.

\begin{aligned}\sigma &= 0 \\ \sigma_{\text{R}} &= \sigma_{\text{I}} = 0\end{aligned}

An example of this was the billiard table example where the ball moved to a new location on the table after being bumped slightly.

# A mathematical description.

For a discussion of stability in fluids we’ll not only have to incorporate the Navier-Stokes equation as we’ve done, but will also have to bring in the heat equation. Unfortunately that isn’t in the scope of this course to derive. Let’s consider as system heated on a bottom plate, and consider the fluid and convection due to heating. This system is illustrated in figure (\ref{fig:continuumL21:continuumL21Fig4})

Figure 8: Fluid in cavity heated on the bottom plate.

\begin{aligned}\rho \frac{\partial {\mathbf{u}}}{\partial {t}} + \rho (\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u} = - \boldsymbol{\nabla} p + \mu \boldsymbol{\nabla}^2 \mathbf{u} - \rho \hat{\mathbf{z}} g.\end{aligned} \hspace{\stretch{1}}(4.3)

For steady state with $\mathbf{u} = 0$ initially (our base state), we’ll call the following the equation of the base state

\begin{aligned}\boxed{\boldsymbol{\nabla} p_s = -\rho_s \hat{\mathbf{z}} g}\end{aligned} \hspace{\stretch{1}}(4.4)

We’ll allow perturbations of each of our variables

\begin{aligned}\mathbf{u} &= \mathbf{u}_\text{base} + \delta \mathbf{u} = 0 + \delta \mathbf{u} \\ p &= p_s + \delta p \\ \rho &= \rho_s + \delta \rho\end{aligned}

After perturbation NS takes the form

\begin{aligned}(\rho_s + \delta \rho )\frac{\partial {(0 + \delta \mathbf{u})}}{\partial {t}} + (\rho_s + \delta \rho) ((0 + \delta \mathbf{u}) \cdot \boldsymbol{\nabla}) (0 + \delta \mathbf{u}) = - \boldsymbol{\nabla} (p_s + \delta p) + \mu \boldsymbol{\nabla}^2 (0 + \delta \mathbf{u}) - (\rho_s + \delta \rho) \hat{\mathbf{z}} g\end{aligned} \hspace{\stretch{1}}(4.5)

Retaining only terms that are of first order of smallness.

\begin{aligned}\rho_s \frac{\partial {\delta \mathbf{u}}}{\partial {t}} = - \boldsymbol{\nabla} p_s - \boldsymbol{\nabla} \delta p + \mu \boldsymbol{\nabla}^2 \delta \mathbf{u} - \rho_s \hat{\mathbf{z}} g - \delta \rho \hat{\mathbf{z}} g\end{aligned} \hspace{\stretch{1}}(4.6)

applying our equation of base state 4.4, we have

\begin{aligned}\rho_s \frac{\partial {\delta \mathbf{u}}}{\partial {t}} = \not{{\rho_s \hat{\mathbf{z}} g}} - \boldsymbol{\nabla} \delta p + \mu \boldsymbol{\nabla}^2 \delta \mathbf{u} - \not{{\rho_s \hat{\mathbf{z}} g}} - \delta \rho \hat{\mathbf{z}} g,\end{aligned} \hspace{\stretch{1}}(4.7)

or

\begin{aligned}\rho_s \frac{\partial {\delta \mathbf{u}}}{\partial {t}} = - \boldsymbol{\nabla} \delta p + \mu \boldsymbol{\nabla}^2 \delta \mathbf{u} - \delta \rho \hat{\mathbf{z}} g.\end{aligned} \hspace{\stretch{1}}(4.8)

we can write

\begin{aligned}\left( \frac{\partial {}}{\partial {t}} - \nu \boldsymbol{\nabla}^2 \right) \delta \mathbf{u} = -\frac{1}{{\rho_s}} \boldsymbol{\nabla} \delta p - \frac{\delta \rho}{\rho_s} \hat{\mathbf{z}} g\end{aligned} \hspace{\stretch{1}}(4.9)

Applying the divergence operation on both sides, and using $\boldsymbol{\nabla} \cdot \mathbf{u} = 0$ so that $\boldsymbol{\nabla} \cdot \delta \mathbf{u} = 0$ we have

\begin{aligned}\left( \frac{\partial {}}{\partial {t}} - \nu \boldsymbol{\nabla}^2 \right) \not{{\boldsymbol{\nabla} \cdot \delta \mathbf{u}}} = -\frac{1}{{\rho_s}} \boldsymbol{\nabla}^2 \delta p - (\hat{\mathbf{z}} \cdot \boldsymbol{\nabla} ) \frac{\delta \rho}{\rho_s} g,\end{aligned} \hspace{\stretch{1}}(4.10)

or

\begin{aligned}\frac{1}{{\rho_s}} \boldsymbol{\nabla}^2 \delta p = - (\hat{\mathbf{z}} \cdot \boldsymbol{\nabla} ) \frac{\delta \rho}{\rho_s} g.\end{aligned} \hspace{\stretch{1}}(4.11)

Assuming that $\rho_s$ is constant (actually that’s already been done above), we can cancel it, leaving

\begin{aligned}\boldsymbol{\nabla}^2 \delta p = - (\hat{\mathbf{z}} \cdot \boldsymbol{\nabla} ) g \delta \rho = -g \frac{\partial {}}{\partial {z}} \delta \rho.\end{aligned} \hspace{\stretch{1}}(4.12)

operating once more with ${\partial {}}/{\partial {z}}$ we have

\begin{aligned}\boldsymbol{\nabla}^2 \frac{\partial {\delta p}}{\partial {z}} = -g \frac{\partial^2 {{\delta \rho}}}{\partial {{z}}^2}.\end{aligned} \hspace{\stretch{1}}(4.13)

Going back to 4.9 and taking only the $z$ component we have

\begin{aligned}\left( \frac{\partial {}}{\partial {t}} - \nu \boldsymbol{\nabla}^2 \right) \delta w = -\frac{1}{{\rho_s}} \frac{\partial {\delta p}}{\partial {z}} - \frac{\delta \rho}{\rho_s} g\end{aligned} \hspace{\stretch{1}}(4.14)

\begin{aligned}\left( \frac{\partial {}}{\partial {t}} - \nu \boldsymbol{\nabla}^2 \right) \boldsymbol{\nabla}^2 \delta w &= -\frac{1}{{\rho_s}} \frac{\partial { \boldsymbol{\nabla}^2 \delta p}}{\partial {z}} - \frac{g}{\rho_s} \boldsymbol{\nabla}^2 \delta \rho \\ &= -\frac{g}{\rho_s} \frac{\partial^2 {{\delta \rho}}}{\partial {{z}}^2} - \frac{g}{\rho_s} \boldsymbol{\nabla}^2 \delta \rho \\ &= -\frac{g}{\rho_s} \left( \frac{\partial^2 {{}}}{\partial {{x}}^2}+\frac{\partial^2 {{}}}{\partial {{y}}^2}\right)\delta \rho \\ &=g \alpha \left( \frac{\partial^2 {{}}}{\partial {{x}}^2}+\frac{\partial^2 {{}}}{\partial {{y}}^2}\right)\delta T\end{aligned}

in the last step we use the following assumed relation for temperature

\begin{aligned}\delta \rho = - \rho_s \alpha \delta T.\end{aligned} \hspace{\stretch{1}}(4.15)

Here $\alpha$ is the coefficient of thermal expansion. This is just a statement that expansion and temperature are related (as we heat something, it expands), with the ratio of the density change relative to the original being linearly related to the change in temperature.

We have finally

\begin{aligned}\left( \frac{\partial {}}{\partial {t}} - \nu \boldsymbol{\nabla}^2 \right) \boldsymbol{\nabla}^2 \delta w = g \alpha \left( \frac{\partial^2 {{}}}{\partial {{x}}^2}+\frac{\partial^2 {{}}}{\partial {{y}}^2}\right)\delta T.\end{aligned} \hspace{\stretch{1}}(4.16)

Solving this is the Rayleigh-Benard instability problem.

While this is a fourth order differential equation, it’s still the same sort of problem logically as we’ve been working on. Our boundary value conditions at $z = 0$ are

\begin{aligned}u, v, w, \delta u, \delta v, \delta w = 0.\end{aligned} \hspace{\stretch{1}}(4.17)

Also relevant will be a similar equation relating temperature and fluid flow rate

\begin{aligned}\left( \frac{\partial {}}{\partial {t}} - \kappa \boldsymbol{\nabla}^2 \right) \delta T = \Delta T \frac{\delta w}{d},\end{aligned} \hspace{\stretch{1}}(4.18)

which we will cover in the next (and final) lecture of the course.