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I was looking into the creeping motion equation: $$ Re \frac{\mathrm{D} \mathbf{u^*}}{\mathrm{D} t^*} = -\nabla^* p^* + \nabla^{*2} \mathbf{u^*} $$

In the case of $ Re << 1$ this equation simplifies to: $$ 0 = -\nabla^* p^* + \nabla^{*2} \mathbf{u^*} $$

It is stated that this equations is independent from time. However, my question is that by definition $\mathbf{u} = \mathbf{u}(\mathbf{x},t)$, so wouldn't there be an implicit dependence from time in the velocity part of the equation? Or is it assumed in the non-dimensionalization that the velocity doesn't depend from time?

Robert Lee
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1 Answers1

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No, there is no time dependence in this any more.
The solution that you get will not depend on t, therefore $\vec u(\vec x,t) = \vec u(\vec x,t+\Delta t)$ for any $\Delta t$.
You can verify this by using a separation of variables ansatz for $z$ and $t$, assuming you're solving for the velocity profile in z-direction.

Note that you're doing the same thing every time you lower the dimensionality of your solution. Your solution then is not really lower-dimensional, it is just the same at all other points.

Another point is that by neglecting the time-derivative via the Reynolds number argument, the equation changes its nature, from hyperbolic to parabolic, so you'd immediately expect a different solution family.

AtmosphericPrisonEscape
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