Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

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}


\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{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)


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{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)


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}


\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{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)


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

\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.

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.


[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.


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