# Peeter Joot's (OLD) Blog.

• 284,968

## Inclined flow without constant height assumption. Setting up the problem, but not actually solving it (reworking vorticity equations)

Posted by peeterjoot on March 20, 2012

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

## Pressure and vorticity equations with the non-linear term retained. (Reworked slightly)

In section 40-2 [1], the identity

\begin{aligned}( \mathbf{u} \cdot \boldsymbol{\nabla} ) \mathbf{u} = (\boldsymbol{\nabla} \times \mathbf{u}) \times \mathbf{u}+ \frac{1}{{2}} \boldsymbol{\nabla} ( \mathbf{u} \cdot \mathbf{u} ),\end{aligned} \hspace{\stretch{1}}(4.37)

is used to put the vorticity equation into a form with one additional portion expressed as a gradient. This is a superior way to handle the inertial term because the curl of that gradient is then killed off.

I’ve expressed the curl as a wedge product, and not a cross product (either works since they related by a constant duality transformation). With the wedge product the identity 4.37 has different signs. That is

\begin{aligned}\mathbf{u} \cdot (\boldsymbol{\nabla} \wedge \mathbf{u})&=(\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u}-\boldsymbol{\nabla}' (\mathbf{u}' \cdot \mathbf{u}) \\ &=(\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u}-\frac{1}{{2}} \boldsymbol{\nabla} (\mathbf{u} \cdot \mathbf{u}),\end{aligned}

Here I’ve used the Hestenes overdot notation [2] to mark the operational range of the gradient $\boldsymbol{\nabla}$ (i.e. indicating that the gradient acts only on one of the $\mathbf{u}$ terms initially). That gives us

\begin{aligned}( \mathbf{u} \cdot \boldsymbol{\nabla} ) \mathbf{u} = \mathbf{u} \cdot (\boldsymbol{\nabla} \wedge \mathbf{u})+\frac{1}{{2}} \boldsymbol{\nabla} (\mathbf{u} \cdot \mathbf{u}).\end{aligned} \hspace{\stretch{1}}(4.38)

Navier-Stokes (for incompressible flows) now takes the form

\begin{aligned}\frac{\partial {\mathbf{u}}}{\partial {t}} + \mathbf{u} \cdot (\boldsymbol{\nabla} \wedge \mathbf{u}) =-\boldsymbol{\nabla} \left( \frac{p}{\rho} + \frac{1}{2} \mathbf{u}^2 + \phi \right) + \nu \boldsymbol{\nabla}^2 \mathbf{u},\end{aligned} \hspace{\stretch{1}}(4.39)

where in our problem we have killed the time dependence and have

\begin{aligned}\phi = -g ( x \sin\alpha, -y \cos\alpha).\end{aligned} \hspace{\stretch{1}}(4.40)

The divergence of $\mathbf{u} \cdot (\boldsymbol{\nabla} \wedge \mathbf{u})$ unfortunately isn’t zero. For exposition purposes, let’s write this out explicitly as a function of the vorticity components

\begin{aligned}\Omega_{rk} = \partial_r u_k - \partial_k u_r.\end{aligned} \hspace{\stretch{1}}(4.41)

Expanding out that divergence we have

\begin{aligned}\boldsymbol{\nabla} \cdot (\mathbf{u} \cdot (\boldsymbol{\nabla} \wedge \mathbf{u})) &=\boldsymbol{\nabla} \cdot \left( u_r \partial_s u_t \mathbf{e}_r \cdot (\mathbf{e}_s \wedge \mathbf{e}_t) \right) \\ &=\boldsymbol{\nabla} \cdot \left( u_r \partial_s u_t (\delta_{rs} \mathbf{e}_t-\delta_{rt} \mathbf{e}_s)\right)\\ &=\partial_k\left( u_r \partial_s u_k \delta_{rs} - u_r \partial_k u_t \delta_{rt} )\right) \\ &=\partial_k \left(u_r\left( \partial_r u_k - \partial_k u_r \right)\right) \end{aligned}

or

\begin{aligned}\boldsymbol{\nabla} \cdot (\mathbf{u} \cdot (\boldsymbol{\nabla} \wedge \mathbf{u})) =\partial_k \left(u_r\Omega_{rk}\right).\end{aligned} \hspace{\stretch{1}}(4.42)

Let’s also, for exposition, expand out the curl of this remaining non-linear term in coordinates. Being a bit smarter this time, we can avoid expressing $\Omega$ in terms of $\boldsymbol{\nabla}$ and $\mathbf{u}$ and leave it as a bivector explicitly. We have

\begin{aligned}\boldsymbol{\nabla} \wedge (\mathbf{u} \cdot \boldsymbol{\Omega})&=\frac{1}{{2}} \boldsymbol{\nabla} \wedge ( u_m \Omega_{ab} \mathbf{e}_m \cdot (\mathbf{e}_a \wedge \mathbf{e}_b) ) \\ &=\frac{1}{{2}} \boldsymbol{\nabla} \wedge ( u_m \Omega_{ab} (\delta_{ma} \mathbf{e}_b - \delta_{mb} \mathbf{e}_a) )\\ &=\frac{1}{{2}} \boldsymbol{\nabla} \wedge ( u_a \Omega_{ab} \mathbf{e}_b-u_b \Omega_{ab} \mathbf{e}_a)\\ &=\boldsymbol{\nabla} \wedge ( u_a \Omega_{ab} \mathbf{e}_b)\end{aligned}

This is

\begin{aligned}\boldsymbol{\nabla} \wedge (\mathbf{u} \cdot \boldsymbol{\Omega}) = \partial_r (u_a \Omega_{ak}) \mathbf{e}_r \wedge \mathbf{e}_k.\end{aligned} \hspace{\stretch{1}}(4.43)

The Navier-Stokes equations are now recast in terms of vorticity as

\begin{subequations}

\begin{aligned}\boldsymbol{\nabla} \cdot (\mathbf{u} \wedge \boldsymbol{\Omega})=-\boldsymbol{\nabla}^2 \left( \frac{p}{\rho} + \frac{1}{2} \mathbf{u}^2 + \phi \right) \end{aligned} \hspace{\stretch{1}}(4.44a)

\begin{aligned}\frac{\partial {\boldsymbol{\Omega}}}{\partial {t}} + \boldsymbol{\nabla} \wedge (\mathbf{u} \cdot \boldsymbol{\Omega}) = \nu \boldsymbol{\nabla}^2 \boldsymbol{\Omega}\end{aligned} \hspace{\stretch{1}}(4.44b)

\begin{aligned}\boldsymbol{\Omega} = \boldsymbol{\nabla} \wedge \mathbf{u}\end{aligned} \hspace{\stretch{1}}(4.44c)

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{u} = 0,\end{aligned} \hspace{\stretch{1}}(4.44d)

\end{subequations}

Having restated things with the $\boldsymbol{\nabla} \mathbf{u}^2$ term moved to the RHS, let’s also now write out 4.44a and 4.44b in coordinate form (we want this for the 2D case). This is

\begin{subequations}

\begin{aligned}\partial_b \left( u_a \Omega_{a b} \right)=-\boldsymbol{\nabla}^2 \left( \frac{p}{\rho} + \frac{1}{2} \mathbf{u}^2 + \phi \right) \end{aligned} \hspace{\stretch{1}}(4.45a)

\begin{aligned}\frac{\partial {\Omega_{mn}}}{\partial {t}} +\partial_m (u_a \Omega_{an}) -\partial_n (u_a \Omega_{am}) =\nu \boldsymbol{\nabla}^2 \Omega_{mn}\end{aligned} \hspace{\stretch{1}}(4.45b)

\end{subequations}

For our problem where we have only $u_1$ and $u_2$ components, and any $\partial_3$ operations are zero, we find

\begin{aligned}\partial_b \left( u_a \Omega_{a b} \right)&=\partial_2 (u_1 \Omega_{12} ) + \partial_1( u_2 \Omega_{21} ) \\ &=(\partial_2 u_1 - \partial_1 u_2 )\Omega_12+ u_1 \partial_2 \Omega_{12} - u_2 \partial_1 \Omega_{12} \\ &=-\Omega_{12}^2 + (u_1 \partial_2 - u_2 \partial_1 ) \Omega_{12} \\ &=-\Omega_{12}^2 - i \cdot (\mathbf{u} \wedge \boldsymbol{\nabla}) \Omega_{12} \\ \end{aligned}

and

\begin{aligned}\partial_1 (u_a \Omega_{a 2}) - \partial_2 (u_a \Omega_{a 1})&=\partial_1 (u_1 \Omega_{1 2}) - \partial_2 (u_2 \Omega_{2 1}) \\ &=\partial_1 (u_1 \Omega_{1 2}) + \partial_2 (u_2 \Omega_{1 2}) \\ &=(\not{{\partial_1 u_1 + \partial_2 u_2}} ) \Omega_{1 2}+ (u_1 \partial_1 + u_2 \partial_2) \Omega_{1 2} \\ &=(\mathbf{u} \cdot \boldsymbol{\nabla}) \Omega_{1 2}\end{aligned}

So we have

\begin{subequations}

\begin{aligned}\Omega_{12}^2 + i \cdot (\mathbf{u} \wedge \boldsymbol{\nabla}) \Omega_{12}=\boldsymbol{\nabla}^2 \left( \frac{p}{\rho} + \frac{1}{2} \mathbf{u}^2 + \phi \right) \end{aligned} \hspace{\stretch{1}}(4.46a)

\begin{aligned}\frac{\partial {\Omega_{12}}}{\partial {t}} +(\mathbf{u} \cdot \boldsymbol{\nabla}) \Omega_{1 2}=\nu \boldsymbol{\nabla}^2 \Omega_{1 2}.\end{aligned} \hspace{\stretch{1}}(4.46b)

\end{subequations}

Here I’ve used $i = \mathbf{e}_1 \mathbf{e}_2$ again, so that the pair of differential operators on the LHS of the respective equations above are

\begin{aligned}i \cdot (\mathbf{u} \wedge \boldsymbol{\nabla}) &= -u_1 \partial_2 + u_2 \partial_1 \\ \mathbf{u} \cdot \boldsymbol{\nabla} &= u_1 \partial_1 + u_2 \partial_2.\end{aligned} \hspace{\stretch{1}}(4.47)

Note that since $\boldsymbol{\nabla}^2 \phi$ could equal zero (as in our problem) we will likely have additional work to ensure that any solution that we find to this set of equations is also still a solution to our original first order Navier-Stokes equation.

# Now what?

The strategy that I’d thought to attempt to tackle this problem, when I’d left it like 4.36 was something along the following lines

\begin{itemize}
\item First ignore the non-linear terms. Find solutions for the homogeneous vorticity and pressure Laplacian equations that satisfy our boundary value conditions, and use that to find a first solution for $h(x)$.
\item Use this to solve for $u$ and $v$ from the vorticity.
\end{itemize}

However, after reworking it using the identity found in Feynman’s dry water chapter, I think it’s best not to try to solve it yet, and study some more first. I have a feeling that there are likely more such techniques that have been developed that will be useful to know before I try to plow my way through things.

Regardless, it’s interesting to see just how tricky the equations of motion become when one doesn’t make unrealistic assumptions. I have a feeling that to actually attempt this specific problem, I may very well need a computer and numerical techniques.

# References

[1] R.P. Feynman, R.B. Leighton, and M.L. Sands. Feynman lectures on physics.[Lectures on physics], chapter The flow of dry water. Addison-Wesley Publishing Company. Reading, Massachusetts, 1963.

[2] D. Hestenes. New Foundations for Classical Mechanics. Kluwer Academic Publishers, 1999.